- Differentiate the constant multiple without needing a formula for \(f\). - A constant term contributes zero to the derivative. - Use the supplied derivative value only after applying the rule.

1. By the constant-multiple and constant rules, \(h'(x)=4f'(x)\). 2. Therefore, \(h'(3)=4\cdot2=8\).

\(h'(3)=8\)

- The derivative of a sum is the sum of the derivatives. - The original function values are not needed for this calculation. - Keep the sign of the second derivative value when combining them.

1. By the sum rule, \(h'(x)=u'(x)+v'(x)\). 2. Thus, \(h'(2)=3+(-5)=-2\).

\(h'(2)=-2\)

- What would the graph of a constant function look like? - A derivative gives the slope at each input.

1. The value of a constant function never changes as \(x\) changes. 2. Its slope is therefore \(0\), so \(f'(x)=0\).

\(f'(x)=0\), because a constant function has zero slope.

- How does multiplying all output values affect tangent slopes? - Pay attention to the signs of both factors.

1. Multiplying a function by a constant multiplies its derivative by the same constant. 2. Therefore, \(g'(2)=-2f'(2)=(-2)(-3)=6\).

\(g'(2)=6\)

- Which earlier derivative rules apply to each term? - How can you put the two results together?

1. Differentiate each term: \((x^2)'=2x\) and \((4x)'=4\). 2. Add them: \(f'(x)=2x+4\).

\(f'(x)=2x+4\)

- How does the derivative of a difference relate to the derivatives of its two parts? - Keep the order of the two functions the same when subtracting their slopes.

1. The difference rule gives \(h'(1)=f'(1)-g'(1)=6-2=4\).

\(h'(1)=4\)

- What is the slope of a constant function? - How does the minus sign affect the derivative of the power term?

1. The derivative of the constant \(8\) is \(0\). 2. The derivative of \(x^3\) is \(3x^2\). Keeping the subtraction gives \(p'(x)=-3x^2\).

\(p'(x)=-3x^2\); the constant term has derivative \(0\).

- What is the derivative rule for a sum? - Differentiate the two functions separately. - Compare the sum of the derivatives with their product. - Evaluate both expressions at a simple input to show they are different.

1. The correct sum rule is \(f'(x)=g'(x)+h'(x)\). 2. For the example, \(g'(x)=3x^2\) and \(h'(x)=2x\). 3. Therefore, the correct derivative is \(f'(x)=3x^2+2x\). 4. Cédric's rule would give \(g'(x)h'(x)=(3x^2)(2x)=6x^3\). 5. These expressions are not equal in general. For example, at \(x=1\), they give \(5\) and \(6\), respectively.

The conjecture is false. The sum rule gives \(f'(x)=g'(x)+h'(x)\). For the example, the correct derivative is \(3x^2+2x\), while the product of the derivatives is \(6x^3\).

- Distinguish multiplication from addition. - What does the constant-multiple rule say? - What happens to a constant coefficient multiplying a power of \(x\)? - Could the graph of \(5x^4\) have slope \(0\) everywhere?

1. Akira is confusing a constant term with a constant factor. The number \(5\) multiplies the variable expression; it is not a separate constant term. 2. By the constant-multiple rule, a constant factor remains: \(\frac{d}{dx}[c\,g(x)]=c\,g'(x)\). 3. Since \(\frac{d}{dx}(x^4)=4x^3\), the derivative is \(f'(x)=5(4x^3)=20x^3\).

Akira incorrectly differentiated a constant factor as though it were a separate constant term. The correct derivative is \(f'(x)=20x^3\).

- First find the point through which the normal line passes. - Tangent and normal slopes are negative reciprocals. - Use the derivative to find the tangent slope. - Substitute the slope and point into a line equation.

1. Find the point: \(f(4)=0.5\cdot4^2-2\cdot4+4=4\), so the point is \((4,4)\). 2. Differentiate: \(f'(x)=x-2\). The tangent slope is \(f'(4)=2\). 3. The normal slope is the negative reciprocal: \(m_n=-\frac{1}{2}\). 4. Use point-slope form: \(y-4=-\frac{1}{2}(x-4)\), so \(y=-0.5x+6\).

\(n: y=-0.5x+6\)

- The derivative at a point gives the tangent line's slope. - How are a line's slope \(m\) and its angle \(\alpha\) related? - Use degree mode when evaluating the inverse tangent.

1. Differentiate: \(f'(x)=-1.5x^2+4x\). 2. Find the tangent slope at \(x=1\): \(m=f'(1)=-1.5(1)^2+4(1)=2.5\). 3. The angle \(\alpha\) satisfies \(\tan(\alpha)=m\), so \(\alpha=\arctan(2.5)\). 4. In degree mode, \(\alpha\approx 68.2^\circ\).

\(68.2^\circ\)

- Evaluate the function to find the point on the graph. - Tangent and normal slopes are negative reciprocals. - Use point-slope form after finding the normal slope.

1. Find the point: \(f(3)=3^2-4\cdot3+1=-2\), so the point is \((3,-2)\). 2. Differentiate: \(f'(x)=2x-4\). The tangent slope is \(f'(3)=2\). 3. The normal slope is \(-\frac{1}{2}\). 4. Use point-slope form: \(y+2=-\frac{1}{2}(x-3)\), so \(y=-0.5x-0.5\).

\(n: y=-0.5x-0.5\)

- Use the derivative to find the tangent slope at the given point. - The normal line is perpendicular to the tangent line. - Perpendicular nonvertical lines have slopes whose product is \(-1\).

1. Find the point: \(f(2)=\frac{1}{4}\cdot2^3+2=4\), so \(P=(2,4)\). 2. Differentiate: \(f'(x)=\frac{3}{4}x^2+1\). The tangent slope is \(f'(2)=4\). 3. The normal slope is \(-\frac{1}{4}\). 4. Use point-slope form: \(y-4=-\frac{1}{4}(x-2)\), so \(y=-0.25x+4.5\).

\(n: y=-0.25x+4.5\)

- Evaluate the function to find the point on the graph. - Use the derivative for the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Write the line using its slope and the point.

1. Find the point: \(f(2)=\frac{1}{2}\cdot2^3-2\cdot2=0\), so the point is \((2,0)\). 2. Differentiate: \(f'(x)=\frac{3}{2}x^2-2\). The tangent slope is \(f'(2)=4\). 3. The normal slope is \(-\frac{1}{4}\). 4. Use point-slope form: \(y=-\frac{1}{4}(x-2)\), so \(y=-0.25x+0.5\).

\(n: y=-0.25x+0.5\)

- Parallel lines have equal slopes. - Set the derivative equal to the slope of the given line. - Find the point of tangency after finding its x-coordinate. - Use point-slope form to write the line.

1. Differentiate: \(f'(x)=0.5x+1\). 2. A parallel tangent must have slope \(2\), so solve \(0.5x+1=2\). This gives \(x=2\). 3. Find the point of tangency: \(f(2)=0.25(2)^2+2-3=0\), so the point is \((2,0)\). 4. Write the tangent line: \(y-0=2(x-2)\), so \(y=2x-4\).

\(y=2x-4\)

- Substitute a point into the function to require that it lies on the graph. - Differentiate while treating the parameter as a constant. - A specified tangent slope gives an equation involving the derivative. - Notice whether the derivative at a given input depends on the parameter.

1. For the graph to pass through \((2,5)\), require \(5=4a-8+1\). Thus, \(4a=12\), so \(a=3\). 2. Differentiate: \(f_a'(x)=2ax-4\). At \(x=0\), \(f_a'(0)=-4\), independent of \(a\). 3. For slope \(10\) at \(x=1\), solve \(f_a'(1)=10\): \(2a-4=10\). Thus, \(a=7\).

a) \(a=3\) b) \(-4\) c) \(a=7\)

- Recall the meaning of a zero exponent. - What happens when an expression is multiplied by \(0\)? - Is \(x^{-1}\) defined at every real value of \(x\)?

1. For \(x\neq0\), \(1=x^0\). 2. Applying the power rule with \(n=0\) gives \(\frac{d}{dx}(x^0)=0x^{-1}=0\) for \(x\neq0\). 3. This agrees with the constant rule. The power-rule expression \(x^{-1}\) is undefined at \(x=0\), so this particular argument does not apply there. However, the original constant function is differentiable at \(0\), and its derivative there is also \(0\).

1) \(f(x)=x^0\) for \(x\neq0\) 2) \(f'(x)=0x^{-1}=0\) for \(x\neq0\) 3) The power-rule argument excludes \(x=0\), but the constant function itself still satisfies \(f'(0)=0\).

- First find the complete point of tangency. - Use the power rule to find the derivative. - Evaluate the derivative at \(x=-2\). - Use point-slope form for the tangent line.

1. Find the point of tangency: \(f(-2)=\frac{1}{4}(-2)^4-(-2)^2=0\), so \(P=(-2,0)\). 2. Differentiate: \(f'(x)=x^3-2x\). 3. Find the slope: \(f'(-2)=-8+4=-4\). 4. Use point-slope form: \(y=-4(x+2)\), so \(y=-4x-8\).

\(y=-4x-8\)

- Rewrite reciprocals and radicals using negative or rational exponents. - A constant factor remains when differentiating. - Apply \(\frac{d}{dx}(x^n)=nx^{n-1}\).

1. For part a), \(f'(x)=2.5(4x^3)=10x^3\). 2. For part b), \(f'(x)=\frac{5}{12}(6x^5)=\frac{5}{2}x^5\). 3. Rewrite part c) as \(f(x)=-3x^{-2}\). Then \(f'(x)=6x^{-3}=\frac{6}{x^3}\). 4. Rewrite part d) as \(f(x)=6x^{1/2}\). Then \(f'(x)=3x^{-1/2}=\frac{3}{\sqrt{x}}\).

a) \(f'(x)=10x^3\) b) \(f'(x)=\frac{5}{2}x^5\) c) \(f'(x)=\frac{6}{x^3}\), where \(x\neq0\) d) \(f'(x)=\frac{3}{\sqrt{x}}\), where \(x>0\)

- Differentiate a sum one term at a time. - Keep constant coefficients when applying the power rule. - What is the derivative of a constant? - Recall that \(x=x^1\).

1. Differentiate \(4x^6\) to get \(24x^5\). 2. Differentiate \(-\frac{1}{2}x^4\) to get \(-2x^3\). 3. The derivative of \(3x\) is \(3\), and the derivative of \(-5\) is \(0\). 4. Therefore, \(f'(x)=24x^5-2x^3+3\).

\(f'(x)=24x^5-2x^3+3\)

- Rewrite the expression so that the term containing \(k\) is isolated. - Find the x-value that makes the coefficient of \(k\) equal to zero. - Evaluate the derivative at that x-value.

1. Rewrite the function as \(f_k(x)=k(x-2)^2+2x-3\). The parameter term is zero when \(x=2\). 2. Then \(f_k(2)=1\), so every graph passes through \(P(2, 1)\). 3. Differentiate: \(f_k'(x)=2kx+2-4k\). At \(x=2\), \(f_k'(2)=4k+2-4k=2\), independent of \(k\). Therefore, every graph has slope \(2\) at \(P\).

a) \(P(2, 1)\) b) Every graph has slope \(2\) at \(P\).

- Use the drawn tangent to make a rough angle estimate before calculating. - The derivative gives the exact tangent slope at the specified input. - Relate slope to inclination with \(m=\tan(\alpha)\).

1. From the graph, the tangent rises about one unit for each unit it moves to the right, so an estimate near \(45^\circ\) is reasonable. 2. Differentiate: \(f'(x)=x\). 3. At \(x=1\), the tangent slope is \(m=f'(1)=1\). 4. Since \(m=\tan(\alpha)\), \(\alpha=\arctan(1)=45^\circ\).

Graphical estimate: about \(45^\circ\); exact angle: \(\alpha=45^\circ\).

- First use the drawn tangent to determine the sign of its slope. - Compute the exact slope from the derivative at \(x=0.5\). - Convert a negative inverse-tangent value to the standard inclination interval \([0^\circ,180^\circ)\).

1. The drawn tangent falls from left to right, so its slope is negative and its angle of inclination is obtuse. 2. Differentiate: \(f'(x)=-2x\). 3. At \(x=0.5\), the tangent slope is \(m=f'(0.5)=-1\). 4. The calculator value \(\arctan(-1)=-45^\circ\) corresponds to an angle of inclination of \(135^\circ\) in \([0^\circ,180^\circ)\).

The angle is obtuse, and its exact measure is \(135^\circ\).

- Differentiate the polynomial term by term. - Evaluate the derivative at the specified input. - Compare the result with the slope of a horizontal line.

1. Differentiate term by term: \(f'(x)=3x^2-6x+3\). 2. Evaluate at \(x=1\): \(f'(1)=0\). 3. Since a horizontal tangent has slope \(0\), the graph has a horizontal tangent at \(x=1\).

\(f'(x)=3x^2-6x+3\) and \(f'(1)=0\), so the tangent is horizontal at \(x=1\).

- Rewrite the reciprocal using a negative exponent. - Differentiate both terms and evaluate at \(x=1\). - Use the original function, not the derivative, for the y-coordinate.

1. Rewrite \(f(x)=x^{-1}+x\). Then \(f'(x)=-x^{-2}+1=-\frac{1}{x^2}+1\). 2. Since \(f'(1)=0\), the tangent is horizontal at \(x=1\). 3. The original function gives \(f(1)=2\), so the point of tangency is \((1,2)\).

\(f'(x)=-\frac{1}{x^2}+1\), \(f'(1)=0\), and the point of tangency is \((1,2)\).

- Evaluate the original function to locate the point of tangency. - Differentiate the polynomial term by term for the tangent slope. - Use point-slope form to write the line.

1. The point of tangency is \((3,-0.5)\). 2. Differentiate term by term: \(f'(x)=x-2\), so the tangent slope is \(f'(3)=1\). 3. Using point-slope form, \(y+0.5=x-3\), so \(y=x-3.5\).

\(t:y=x-3.5\)

- Apply the Power Rule to the quadratic term. - Keep the constant factor attached to each differentiated term. - Evaluate only after finding the derivative function.

1. Differentiate term by term: \(f'(x)=4x-3\). 2. Evaluate the derivative: \(f'(1)=1\). 3. The instantaneous rate of change at \(x=1\) is \(1\).

\(f'(1)=1\)

- Differentiate each term separately. - The derivative of the constant term is \(0\). - Substitute \(x=2\) only after differentiating.

1. Differentiate term by term: \(g'(x)=3x^2-8x\). 2. Evaluate at \(x=2\): \(g'(2)=12-16=-4\). 3. The graph has slope \(-4\) at \(x=2\).

\(g'(2)=-4\)

- Differentiate each term while treating \(a\) as a constant coefficient. - Use the derivative value at the specified input to create an equation for \(a\). - Solve the resulting linear equation only after differentiation.

1. Differentiate using the constant-multiple and sum rules: \(p'(x)=3ax^2+2\). 2. Substitute \(x=1\): \(3a+2=11\). 3. Solving gives \(a=3\).

\(a=3\)

- Simplify the expression before applying differentiation rules. - Keep track of the minus sign outside the parentheses. - What is the derivative of the constant term?

1. Distribute the subtraction: \(f(x)=5-2x^3+3x\). 2. Differentiate each term: \(f'(x)=0-6x^2+3=-6x^2+3\).

\(f'(x)=-6x^2+3\)

- For one method, combine like powers before differentiating. - For the other, keep the outer constant multiplier attached to the parentheses. - Compare the simplified derivative expressions at the end.

1. Simplify: \(f(x)=2x^2+6x-3x^2=-x^2+6x\). Thus, \(f'(x)=-2x+6\). 2. Alternatively, differentiate the unsimplified expression: \(f'(x)=2(2x+3)-6x=4x+6-6x=-2x+6\). 3. Both methods agree.

Simplify first: \(f(x)=-x^2+6x\), so \(f'(x)=-2x+6\). Differentiate first: \(f'(x)=2(2x+3)-6x=-2x+6\).

- First find the tangent slope at the requested input. - How are the slopes of two perpendicular nonvertical lines related?

1. Differentiate term by term: \(f'(x)=2x+1\). 2. The tangent slope at \(x=1\) is \(3\). 3. The normal slope is the negative reciprocal, \(-\frac{1}{3}\).

Normal slope: \(-\frac{1}{3}\).

- A line equation needs a point and a slope. - Find the tangent slope before thinking about a perpendicular line. - Check that your answer passes through the point on the curve.

1. Differentiate: \(f'(x)=2x-3\). At \(x=0\) the tangent slope is \(-3\). 2. The normal slope is \(\frac{1}{3}\). The point on the graph is \((0,2)\). 3. The normal line is \(y=\frac{1}{3}x+2\).

\(y=\frac{1}{3}x+2\)

- Connect the tangent slope to a familiar trigonometric ratio. - Is the slope positive or negative at the given input?

1. Differentiate: \(f'(x)=x\), so the tangent slope at \(1\) is \(1\). 2. The angle \(\alpha\) satisfies \(\tan\alpha=1\), and the slope is positive. Therefore, \(\alpha=45^\circ\).

\(45^\circ\)

- Use the derivative to find the tangent slope. - Think about how the tangent's direction relates to a counterclockwise angle. - Check the requested angle against the slope you calculated.

1. Differentiate: \(f'(x)=-2x\), so the tangent slope at \(1\) is \(-2\). 2. A negative slope gives an inclination between \(90^\circ\) and \(180^\circ\). Thus, \(\alpha=180^\circ-\arctan2\approx116.6^\circ\).

\(116.6^\circ\)

- Treat \(k\) as a constant when differentiating with respect to \(x\). - Evaluate the derivative at each input before deciding what changes with \(k\).

1. Differentiate term by term: \(f_k'(x)=2kx+2\). 2. At \(x=0\), the slope is \(2\). At \(x=2\), it is \(4k+2\). 3. Only the slope at \(x=0\) is independent of \(k\).

\(f_k'(0)=2\); \(f_k'(2)=4k+2\). The slope at \(x=0\) does not depend on \(k\).

- What do parallel nonvertical lines have in common? - Write the tangent slope in terms of \(a\). - Check the condition using the value you find.

1. The given line has slope \(7\), so the tangent must also have slope \(7\). 2. Differentiate: \(f_a'(x)=2ax+3\). At \(x=1\) the slope is \(2a+3\). 3. Solve \(2a+3=7\), giving \(a=2\).

\(a=2\)

- Differentiate the polynomial term by term. - A horizontal tangent corresponds to a derivative value of \(0\). - For a prescribed slope, set the derivative equal to that slope. - Substitute each resulting x-coordinate into the original function to find the point.

1. Differentiate using the power, constant, and sum rules: \(f'(x)=-2x+4\). 2. A horizontal tangent has slope \(0\), so solve \(-2x_0+4=0\). This gives \(x_0=2\). 3. Evaluate the function: \(f(2)=2\). Therefore, \(P=(2,2)\). 4. For slope \(6\), solve \(-2q+4=6\). This gives \(q=-1\). 5. Evaluate the function: \(f(-1)=-7\). Therefore, \(Q=(-1,-7)\).

a) \(f'(x)=-2x+4\) b) \(P=(2,2)\) c) \(Q=(-1,-7)\)

- What must be true about the slopes of parallel lines? - Read the slope of the given line from its equation. - Use the slope and point of tangency to write the tangent line equation.

1. Differentiate: \(g'(x)=x-3\). 2. Parallel lines have equal slopes. The given line has slope \(2\), so solve \(g'(b)=2\): \(b-3=2\), giving \(b=5\). 3. Evaluate the function: \(g(5)=\frac{1}{2}\cdot 5^2-3\cdot 5=-2.5\). Thus, \(B=(5,-2.5)\). 4. Use slope \(2\) and point \((5,-2.5)\): \(y+2.5=2(x-5)\). 5. Simplify: \(y=2x-12.5\).

a) \(g'(x)=x-3\) b) \(B=(5,-2.5)\) c) \(y=2x-12.5\)

- Begin by differentiating the original polynomial rather than using a supplied derivative. - Set the derivative equal to each required slope. - Parallel lines have equal slopes. - After finding an input, use the original function to recover the corresponding point.

1. Differentiate term by term: \(f'(x)=3x^2-12\). 2. Evaluate the derivative: \(f'(-1)=-9\) and \(f'(3)=15\). 3. For slope \(15\), solve \(3x^2-12=15\). Then \(x^2=9\), so \(x=-3\) or \(x=3\). 4. A line parallel to \(y=-9x+2\) must have slope \(-9\). Solving \(3x^2-12=-9\) gives \(x=-1\) or \(x=1\). Since \(f(-1)=16\) and \(f(1)=-6\), the points are \((-1,16)\) and \((1,-6)\). 5. At \(x=0\), \(f(0)=5\) and \(f'(0)=-12\). The tangent line is \(y=-12x+5\).

a) \(f'(-1)=-9\) and \(f'(3)=15\) b) \(x=-3\) and \(x=3\) c) \((-1,16)\) and \((1,-6)\) d) \(y=-12x+5\)

- Differentiate the original polynomial term by term first. - Set the derivative equal to the requested slope and solve. - Read the comparison line's slope from its equation. - Substitute each relevant input into the original function to find the point.

1. Differentiate term by term: \(g'(x)=x^2+2x-3\). 2. Evaluate: \(g'(-2)=-3\) and \(g'(1)=0\). 3. For slope \(5\), solve \(x^2+2x-3=5\). Factoring \(x^2+2x-8=(x+4)(x-2)\) gives \(x=-4\) or \(x=2\). 4. A tangent parallel to \(y=-3x+10\) must have slope \(-3\). Solve \(x^2+2x-3=-3\), giving \(x=0\) or \(x=-2\). 5. Since \(g(0)=0\) and \(g(-2)=\frac{22}{3}\), the points are \((0,0)\) and \((-2,\frac{22}{3})\).

a) \(g'(-2)=-3\) and \(g'(1)=0\) b) \(x=-4\) and \(x=2\) c) \((0,0)\) and \((-2,\frac{22}{3})\)

- Find the point of tangency and the derivative at \(x=3\). - A line is determined by a point and its slope. - The derivative gives the tangent line's slope. - What is the y-coordinate of every point on the x-axis?

1. Find the point of tangency: \(f(3)=\frac{1}{2}\cdot 3^2-2\cdot 3+1=-0.5\), so \(P=(3,-0.5)\). 2. Differentiate: \(f'(x)=x-2\). 3. Find the tangent slope: \(f'(3)=1\). 4. Write the tangent line: \(y+0.5=x-3\), so \(y=x-3.5\). 5. At an x-intercept, \(y=0\). Solve \(0=x-3.5\) to get \(x=3.5\).

\((3.5,0)\)

- First find the point where the tangent touches the graph. - Use the derivative to find the tangent slope. - Write the tangent line equation. - A point on the x-axis has y-coordinate \(0\).

1. Find the point of tangency: \(g(1)=1^3-4(1)=-3\), so the point is \((1,-3)\). 2. Differentiate: \(g'(x)=3x^2-4\). 3. Find the tangent slope: \(g'(1)=3(1)^2-4=-1\). 4. Write the tangent line: \(y+3=-(x-1)\), so \(y=-x-2\). 5. Set \(y=0\) to find the x-intercept: \(0=-x-2\), giving \(x=-2\).

\((-2,0)\)

- Find the slope of a line whose angle of inclination is \(135^\circ\). - Express the tangent slope at an unknown input using the derivative. - Set the derivative equal to the required slope and solve.

1. Convert the angle to a slope: \(m=\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=\frac{1}{2}x+1\). 3. Set the derivative equal to the required slope: \(\frac{1}{2}x_0+1=-1\). 4. Solve: \(\frac{1}{2}x_0=-2\), so \(x_0=-4\).

\(x_0=-4\)

- A tangent equation needs both a point on the curve and a slope there. - Check whether the line you found passes through the required point. - Is your final angle consistent with whether the tangent rises or falls to the right?

1. Find the point of tangency: \(f(3)=\frac{1}{3}\cdot3^3-3^2-2\cdot3=-6\), so the point is \((3,-6)\). 2. Differentiate: \(f'(x)=x^2-2x-2\). 3. Find the slope: \(f'(3)=3^2-2\cdot3-2=1\). 4. Use point-slope form: \(y+6=x-3\), so \(y=x-9\). 5. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

The tangent line is \(y=x-9\), and its angle with the positive x-axis is \(45^\circ\).

- Convert the given angle to a slope. - Set the derivative equal to that slope to find the x-coordinate. - Substitute the x-coordinate into the original function, then write the tangent line.

1. Convert the angle to a slope: \(m=\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=2x-4\). 3. Set the derivative equal to the tangent slope: \(2x-4=-1\). Solving gives \(x=1.5\). 4. Find the y-coordinate: \(f(1.5)=(1.5)^2-4(1.5)+2=-1.75\). Thus, \(B=(1.5,-1.75)\). 5. Use point-slope form: \(y+1.75=-(x-1.5)\), so \(y=-x-0.25\).

\(B=(1.5,-1.75)\) and \(y=-x-0.25\)

- Before writing either line, which point must both lines pass through? - The tangent and normal are perpendicular. What condition does that place on their slopes? - Angle of inclination is measured counterclockwise from the positive x-axis, even for a negative slope.

1. Find the point: \(f(4)=0.5\cdot4^2-2\cdot4=0\), so the point is \((4,0)\). 2. Differentiate: \(f'(x)=x-2\). The tangent slope is \(m_t=f'(4)=2\). 3. The tangent line is \(y=2(x-4)=2x-8\). 4. The normal slope is \(m_n=-\frac{1}{2}\), so the normal line is \(y=-0.5(x-4)=-0.5x+2\). 5. The tangent angle is \(\alpha_t=\arctan(2)\approx63.43^\circ\). The normal angle is \(\alpha_n=\arctan(-0.5)+180^\circ\approx153.43^\circ\).

\(t:y=2x-8\) \(n:y=-0.5x+2\) \(\alpha_t\approx63.43^\circ\), \(\alpha_n\approx153.43^\circ\)

- The tangent slope is the derivative at the point. - The slopes of a tangent and its normal are negative reciprocals when both are defined. - Convert an angle of inclination to a slope using tangent. - To verify a tangent line, check both the point and the slope.

1. Differentiate: \(f'(x)=x-3\). 2. At \(x=4\), \(f(4)=-3\) and \(f'(4)=1\). The tangent line is \(y+3=x-4\), or \(y=x-7\). The normal slope is \(-1\), so the normal line is \(y+3=-(x-4)\), or \(y=-x+1\). 3. An angle of \(135^\circ\) corresponds to slope \(\tan(135^\circ)=-1\). Solve \(x-3=-1\) to get \(x=2\). 4. At \(x=2\), \(f(2)=-3\) and \(f'(2)=-1\). The tangent line is \(y+3=-(x-2)\), which simplifies to \(y=-x-1\). Therefore, the given line is tangent at \(x=2\).

a) Tangent: \(y=x-7\); normal: \(y=-x+1\) b) \(x=2\) c) Yes. At \(x=2\), the point is \((2,-3)\) and the tangent slope is \(-1\), giving \(y=-x-1\).

- A line parallel to the x-axis has slope \(0\). - Substitute the given input into the original function to find the point. - A normal line has slope equal to the negative reciprocal of the tangent slope. - Use point-slope form to write each line.

1. Differentiate: \(f'(x)=x^2-1\). 2. At \(x=2\), \(f(2)=\frac{2}{3}\) and \(f'(2)=3\). The tangent line is \(y-\frac{2}{3}=3(x-2)\), or \(y=3x-\frac{16}{3}\). 3. A tangent parallel to the x-axis has slope \(0\). Solve \(x^2-1=0\), giving \(x=-1\) and \(x=1\). 4. At \(x=0\), \(f(0)=0\) and the tangent slope is \(-1\). The normal slope is \(1\), so the normal line is \(y=x\).

a) \(y=3x-\frac{16}{3}\) b) \(x=-1\) and \(x=1\) c) \(y=x\)

- First verify that the two functions have the same value at \(x=1\). - Then compare their derivative values at that input. - A common tangent must share both the point and the slope.

1. The function values agree: \(f(1)=3\) and \(g(1)=3\), so both graphs pass through \((1,3)\). 2. Differentiate term by term: \(f'(x)=2x+1\) and \(g'(x)=-2x+5\). 3. The slopes also agree at \(x=1\): \(f'(1)=g'(1)=3\). Therefore, the graphs are tangent at \((1,3)\). 4. The common tangent line is \(y-3=3(x-1)\), so \(y=3x\).

1. Both graphs pass through \((1,3)\), and \(f'(1)=g'(1)=3\), so they are tangent there. 2. \(y=3x\)

- The derivative gives the graph's slope at a point. - A tangent line with slope \(0\) is horizontal. - The normal slope is the negative reciprocal of the tangent slope. - Use point-slope form for each line.

1. Differentiate: \(f'(x)=x^3-4x\). 2. At \(x=1\), \(f'(1)=1-4=-3\). 3. At \(x=2\), \(f(2)=1\) and \(f'(2)=0\). The tangent line is horizontal, so its equation is \(y=1\). 4. At \(x=-1\), \(f(-1)=\frac{13}{4}\) and the tangent slope is \(f'(-1)=3\). The normal slope is \(-\frac{1}{3}\). 5. The normal line is \(y-\frac{13}{4}=-\frac{1}{3}(x+1)\), which simplifies to \(y=-\frac{1}{3}x+\frac{35}{12}\).

a) \(-3\) b) \(y=1\) c) \(y=-\frac{1}{3}x+\frac{35}{12}\)

- Convert the known normal slope to the tangent slope. - The derivative gives the tangent slope at each input. - Set the derivative equal to the required tangent slope. - Evaluate the function before writing the line equation.

1. The tangent slope is the negative reciprocal of the normal slope: \(m_t=-\frac{1}{0.5}=-2\). 2. Differentiate: \(f'(x)=2x-4\). Set \(f'(x_0)=-2\): \(2x_0-4=-2\), so \(x_0=1\). 3. Evaluate the function: \(f(1)=2\), so the point is \((1,2)\). 4. The normal line is \(y-2=0.5(x-1)\), or \(y=0.5x+1.5\).

a) \(x_0=1\) b) \(n: y=0.5x+1.5\)

- Find the tangent slope at \(A\) using the derivative. - Convert the tangent slope to the normal slope. - Write the normal line through \(A\). - Solve the two line equations as a system.

1. Differentiate: \(p'(x)=2x-4\). The tangent slope at \(x=3\) is \(p'(3)=2\). 2. The normal slope is \(-\frac{1}{2}\). 3. Through \(A(3,3)\), the normal line is \(y-3=-0.5(x-3)\), or \(y=-0.5x+4.5\). 4. Set the normal equation equal to \(y=x-3\): \(-0.5x+4.5=x-3\). Thus, \(7.5=1.5x\), so \(x=5\). 5. Then \(y=5-3=2\), so \(S=(5,2)\).

\(S=(5,2)\)

- Set the derivative equal to the given slope. - Substitute each solution into the original function to find the points. - Use a coordinate-area formula for a triangle whose one vertex is the origin. - Take the absolute value so the area is positive.

1. Differentiate: \(f'(x)=x^2-2x-2\). 2. Set the derivative equal to \(1\): \(x^2-2x-2=1\), or \(x^2-2x-3=0\). Thus, \(x=-1\) or \(x=3\). 3. Find the corresponding points: \(f(-1)=\frac{2}{3}\) and \(f(3)=-6\). Thus, the points are \((-1,\frac{2}{3})\) and \((3,-6)\). 4. With one vertex at the origin, the area is \(\frac{1}{2}|x_1y_2-x_2y_1|\). 5. Therefore, \(A=\frac{1}{2}|(-1)(-6)-3(\frac{2}{3})|=\frac{1}{2}|6-2|=2\).

\(2\) square units

- The y-intercept occurs where \(x=0\). - A horizontal tangent has derivative \(0\). - The derivative at a point gives the tangent slope. - Use point-slope form to write each tangent line.

1. The y-intercept occurs at \(x=0\), and \(f(0)=11\). Differentiate: \(f'(x)=3x^2-6x-9\), so \(f'(0)=-9\). The tangent line is \(y-11=-9(x-0)\), or \(y=-9x+11\). 2. A horizontal tangent has slope \(0\). Solve \(3x^2-6x-9=0\). Dividing by \(3\) gives \(x^2-2x-3=0\), so \(x=-1\) or \(x=3\). 3. At \(x=4\), \(f(4)=-9\) and \(f'(4)=15\). Thus, \(P=(4,-9)\), and the tangent line is \(y+9=15(x-4)\), or \(y=15x-69\).

1. At \(x=0\), \(f(0)=11\) and \(f'(0)=-9\), so the tangent is \(y=-9x+11\). 2. \(x=-1\) and \(x=3\) 3. \(P=(4,-9)\) and \(y=15x-69\)

- Differentiate the polynomial term by term to obtain the tangent slope. - Use the original function to locate the point of tangency. - Write the line with point-slope form. - Relate slope to inclination with the tangent function and use the requested precision.

1. Find the point of tangency: \(f(2)=1\), so \(P=(2,1)\). 2. Differentiate term by term: \(f'(x)=3x^2-4x\). 3. The tangent slope is \(f'(2)=4\). 4. The tangent line is \(y-1=4(x-2)\), or \(y=4x-7\). 5. The angle \(\alpha\) satisfies \(\tan(\alpha)=4\), so \(\alpha=\arctan(4)\approx75.96^\circ\).

Tangent line: \(y=4x-7\) Angle: \(\alpha\approx75.96^\circ\)

- Use the derivative to find the tangent slope. - Substitute \(x=1\) into the original function to find the point. - Relate the slope to the angle using tangent. - Write the line in slope-intercept or point-slope form.

1. Find the point of tangency: \(f(1)=-0.25(1)^4+1^2+1=1.75\), so \(P=(1,1.75)\). 2. Differentiate: \(f'(x)=-x^3+2x\). 3. Find the slope: \(f'(1)=1\). 4. Write the tangent line: \(y-1.75=x-1\), so \(y=x+0.75\). 5. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

Tangent line: \(y=x+0.75\) Angle: \(45^\circ\)

- Parallel lines have equal slopes. - Set the derivative equal to the specified slope. - Find the corresponding points on the graph. - Use each point and the common slope to write a tangent line.

1. Differentiate: \(f'(x)=x^2-4\). 2. The given line has slope \(5\). Set \(f'(x)=5\): \(x^2-4=5\), so \(x=-3\) or \(x=3\). 3. Evaluate the function: \(f(-3)=5\) and \(f(3)=-1\). 4. At \((-3,5)\), the tangent line is \(y-5=5(x+3)\), or \(y=5x+20\). 5. At \((3,-1)\), the tangent line is \(y+1=5(x-3)\), or \(y=5x-16\).

\(y=5x+20\) and \(y=5x-16\)

- Treat the parameter as a constant when differentiating. - Set the derivative at \(x=2\) equal to the required slope. - Find the point of tangency using the corresponding function. - Use point-slope form for the tangent line.

1. Differentiate while treating \(k\) as a constant: \(g_k'(x)=3x^2-2kx\). 2. Require slope \(4\) at \(x=2\): \(3(2)^2-2k(2)=4\). Thus, \(12-4k=4\), so \(k=2\). 3. For \(k=2\), \(g_2(2)=2^3-2(2)^2+4=4\). The point of tangency is \((2,4)\). 4. Using slope \(4\), the tangent line is \(y-4=4(x-2)\), or \(y=4x-4\).

a) \(g_k'(x)=3x^2-2kx\) b) \(k=2\) c) \(y=4x-4\)

- Use the derivative to find the tangent slope. - Find the complete point of tangency. - Use point-slope form to write the line. - Set \(x=0\) to find the y-intercept.

1. Differentiate: \(f'(x)=x-2\). 2. At \(x=-1\), \(f(-1)=1.5\) and \(f'(-1)=-3\). 3. Write the tangent line: \(y-1.5=-3(x+1)\), so \(y=-3x-1.5\). 4. At the y-intercept, \(x=0\). Thus, \(y=-1.5\), and the intercept is \((0,-1.5)\).

Tangent line: \(y=-3x-1.5\) Y-intercept: \((0,-1.5)\)

- Evaluate the function to find the full point. - Use the derivative for the tangent slope. - Perpendicular line slopes are negative reciprocals. - Use point-slope form for both lines.

1. Differentiate: \(g'(x)=x^2-2x\). 2. Evaluate the function and derivative: \(g(3)=2\) and \(g'(3)=3\). 3. The tangent line is \(y-2=3(x-3)\), or \(y=3x-7\). 4. The normal slope is \(-\frac{1}{3}\). Thus, the normal line is \(y-2=-\frac{1}{3}(x-3)\), or \(y=-\frac{1}{3}x+3\).

a) \(t: y=3x-7\) b) \(n: y=-\frac{1}{3}x+3\)

- Substitute a point into the original function to find a parameter. - The derivative gives the slope at a specified x-value. - Apply the power rule term by term. - The origin has x-coordinate \(0\).

1. Substitute \((2, 8)\): \(8-4k+4=8\). Solving gives \(k=1\). 2. Differentiate: \(f_k'(x)=3x^2-2kx+2\). 3. At the origin, \(f_k'(0)=2\), independent of \(k\). 4. For slope \(11\) at \(x=3\), solve \(f_k'(3)=11\): \(27-6k+2=11\). 5. This gives \(6k=18\), so \(k=3\).

a) \(k=1\) b) The slope is \(2\). c) \(k=3\)

- Substitute the given point into the function. - Differentiate to find a slope formula. - Evaluate the derivative at the specified x-value. - Decide whether the requested slope is constant or depends on the parameter.

1. Substitute \((1, 2)\): \(1-3+t=2\), so \(t=4\). 2. Differentiate: \(g_t'(x)=3x^2-6x+t\). 3. At the origin, \(g_t'(0)=t\). 4. For slope \(5\) at \(x=2\), solve \(g_t'(2)=5\): \(12-12+t=5\). 5. Therefore, \(t=5\).

a) \(t=4\) b) The slope is \(t\). c) \(t=5\)

- Convert each angle to a slope using tangent. - Set the derivative equal to the required slope. - Check whether the resulting squared equation has real solutions. - Consider the range of \(g'(x)=x^2-1\).

1. An angle of \(45^\circ\) corresponds to slope \(\tan(45^\circ)=1\). Differentiate: \(g'(x)=x^2-1\). 2. Solve \(x^2-1=1\). Then \(x^2=2\), so \(x=\pm\sqrt{2}\). 3. Evaluate the function: \(g(\sqrt{2})=-\frac{\sqrt{2}}{3}\) and \(g(-\sqrt{2})=\frac{\sqrt{2}}{3}\). The points are \((\sqrt{2},-\frac{\sqrt{2}}{3})\) and \((-\sqrt{2},\frac{\sqrt{2}}{3})\). 4. An angle of \(120^\circ\) corresponds to slope \(\tan(120^\circ)=-\sqrt{3}\). The equation \(x^2-1=-\sqrt{3}\) becomes \(x^2=1-\sqrt{3}\). 5. Since \(1-\sqrt{3}<0\), there is no real solution. Equivalently, \(g'(x)=x^2-1\ge -1\), so it can never equal \(-\sqrt{3}\).

1. \((\sqrt{2},-\frac{\sqrt{2}}{3})\) and \((-\sqrt{2},\frac{\sqrt{2}}{3})\) 2. No such point exists because \(g'(x)\ge -1\), while \(\tan(120^\circ)=-\sqrt{3}<-1\).

- A horizontal tangent has slope \(0\). - Treat the parameter \(k\) as a constant when differentiating with respect to \(x\). - Substitute the specified input into the derivative before solving for \(k\).

1. Treat \(k\) as a constant and differentiate: \(f_k'(x)=3kx^2-6x\). 2. A horizontal tangent has slope \(0\), so require \(f_k'(1)=0\). 3. Substitute \(x=1\): \(3k-6=0\). 4. Solving gives \(k=2\).

a) \(f_k'(x)=3kx^2-6x\) b) \(k=2\)

- A horizontal tangent has slope \(0\). - Evaluate the derivative at both specified inputs. - Compare absolute values when comparing steepness.

1. Differentiate: \(g'(x)=x^2-4x+3\). 2. Horizontal tangent lines occur where \(g'(x)=0\). Factor \(x^2-4x+3=(x-1)(x-3)\), so \(x=1\) and \(x=3\). 3. Evaluate the slopes: \(g'(0)=3\) and \(g'(4)=3\). 4. Steepness is determined by the absolute value of slope. Since both slopes have magnitude \(3\), the graph is equally steep at the two points.

a) \(x=1\) and \(x=3\) b) The graph is equally steep at \(x=0\) and \(x=4\); the slope is \(3\) at each point.

- Find the point and the derivative at \(x=1\) in terms of \(k\). - Write the tangent line in point-slope form. - Substitute the coordinates of \(Q\) into the tangent-line equation. - Solve the resulting equation for \(k\).

1. At \(x=1\), the point on the graph is \(B(1, f_k(1))=(1, 5-k)\). 2. Differentiate: \(f_k'(x)=4x^3-2kx\), so the tangent slope at \(x=1\) is \(4-2k\). 3. The tangent line is \(y-(5-k)=(4-2k)(x-1)\). 4. Because the line passes through \(Q(0, 1)\), \(1-(5-k)=(4-2k)(-1)\). 5. Simplifying gives \(k-4=-4+2k\), so \(k=0\).

\(k=0\)

- Rewrite the square root as a power of \(s\). - What are the units of speed divided by distance? - For part c), see what happens to the positive fraction inside the square root when \(s\) gets larger.

1. Rewrite \(v(s)=\sqrt{2g}\,s^{\frac{1}{2}}\). By the power rule, \(v'(s)=\frac{\sqrt{2g}}{2\sqrt{s}}=\sqrt{\frac{g}{2s}}\), for \(s>0\). 2. At \(s=2\), \(v'(2)=\sqrt{\frac{9.81}{4}}\approx1.566\,\text{s}^{-1}\). At \(s=8\), \(v'(8)=\sqrt{\frac{9.81}{16}}\approx0.783\,\text{s}^{-1}\). 3. The derivative gives the change in speed (meters per second) per meter fallen, so its units are \(\text{s}^{-1}\). 4. For \(s>0\), the denominator \(2s\) in \(\frac{g}{2s}\) grows as \(s\) grows. Because \(g>0\) and the square-root function is increasing, \(\sqrt{\frac{g}{2s}}\) decreases as \(s\) increases. No second derivative is needed.

a) \(v'(s)=\sqrt{\frac{g}{2s}}\) for \(s>0\) b) \(v'(2)\approx1.566\,\text{s}^{-1}\), \(v'(8)\approx0.783\,\text{s}^{-1}\). This is the speed gained per extra meter fallen. c) \(\frac{g}{2s}\) decreases as \(s\) increases, so its square root \(v'(s)\) also decreases.

- Rewrite the quadratic in vertex form. - Substitute the point's x-coordinate into the function. - Use the derivative to verify the tangent slope at \(P\). - Use the point and slope to write the tangent line. - Test the origin by substituting \((0,0)\) into the line equation.

1. Complete the square: \(f(x)=(x-2)^2+1\). Therefore, the vertex is \((2,1)\). 2. Evaluate the function: \(f(3)=9-12+5=2\). Thus, \(P=(3,2)\) lies on the graph. 3. Differentiate: \(f'(x)=2x-4\). Then \(f'(3)=2\), so the tangent slope at \(P\) is \(2\). 4. A line with slope \(2\) through \((3,2)\) satisfies \(y-2=2(x-3)\), so \(y=2x-4\). 5. At \(x=0\), the line has value \(-4\), not \(0\). Therefore, it does not pass through the origin.

a) \((2,1)\) b) Yes, because \(f(3)=2\). c) \(f'(3)=2\), so the tangent line is \(y=2x-4\); it does not pass through the origin.

- The derivative gives the tangent slope. - A horizontal tangent has slope \(0\). - Write the general tangent line at \(x=u\). - Expand and combine like terms carefully.

1. Differentiate: \(f'(x)=3x^2-3\). 2. At \(x=2\), \(f(2)=2\) and \(f'(2)=9\). Thus, \(y-2=9(x-2)\), so \(y=9x-16\). 3. A horizontal tangent has slope \(0\). Solve \(3x^2-3=0\), giving \(x=-1\) and \(x=1\). 4. At an arbitrary input \(u\), the point is \((u,u^3-3u)\) and the slope is \(3u^2-3\). 5. Substitute into point-slope form: \(y=(3u^2-3)(x-u)+u^3-3u\). Expanding gives \(y=(3u^2-3)x-2u^3\).

a) \(y=9x-16\) b) \(x=-1\) and \(x=1\) c) Substituting \(f(u)=u^3-3u\) and \(f'(u)=3u^2-3\) into point-slope form simplifies to \(y=(3u^2-3)x-2u^3\).

- What two pieces of information determine a tangent line at an unknown input? - How can both pieces be written in terms of \(a\)? - Check your formula by substituting one specific value for \(a\).

1. Differentiate: \(f'(x)=4x-5\). 2. At \(x=a\), the tangent slope is \(f'(a)=4a-5\), and the point is \((a,2a^2-5a)\). 3. Use point-slope form: \(y=(4a-5)(x-a)+2a^2-5a\). 4. Expand and simplify: \(y=(4a-5)x-4a^2+5a+2a^2-5a=(4a-5)x-2a^2\).

\(t_a:y=(4a-5)x-2a^2\)

- Rewrite the reciprocal term using a negative exponent. - The derivative gives the tangent slope. - Use the point and slope to write the tangent line.

1. Find the point of tangency: \(f(1)=3\), so \(P=(1,3)\). 2. Rewrite \(f(x)=2x^{-1}+x\) and differentiate: \(f'(x)=-2x^{-2}+1=-\frac{2}{x^2}+1\). 3. Find the slope: \(f'(1)=-1\). 4. Use point-slope form: \(y-3=-(x-1)\), so \(y=-x+4\).

\(y=-x+4\)

- Find both tangent slopes using the derivative. - What slope relationship characterizes perpendicular lines? - You may also compare the lines' angles of inclination.

1. Differentiate: \(g'(x)=2x-3\). 2. The tangent slopes are \(g'(1)=-1\) and \(g'(2)=1\). 3. Since the product of the slopes is \((-1)(1)=-1\), the tangent lines are perpendicular. 4. Therefore, their angle of intersection is \(90^\circ\).

\(90^\circ\)

- Parallel lines have equal slopes. - The derivative gives the tangent slope at \(a\). - Set the derivative equal to \(2\) and solve. - Check for more than one solution.

1. Differentiate: \(f'(x)=x^2+2x-6\). 2. The given line has slope \(2\), so require \(f'(a)=2\). 3. Solve \(a^2+2a-6=2\), or \(a^2+2a-8=0\). 4. Factor: \((a+4)(a-2)=0\). Thus, \(a=-4\) or \(a=2\).

\(a=-4\) and \(a=2\)

- How is a tangent's angle connected to its signed slope? - Where on the curve does the derivative match that slope? - Once the input is known, how do you locate the point on the graph?

1. An angle of \(135^\circ\) corresponds to slope \(\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=2x-5\). 3. Set the derivative equal to the required slope: \(2x-5=-1\). Thus, \(x=2\). 4. Evaluate the function: \(f(2)=4-10+2=-4\). Therefore, the point is \((2,-4)\).

\((2,-4)\)

- The derivative represents the tangent slope. - Convert the angle to a slope. - Solve the derivative equation for all possible inputs. - Substitute each input into the original function.

1. An angle of \(45^\circ\) corresponds to slope \(1\). 2. Differentiate: \(f'(x)=x^2-x-1\). 3. Set the derivative equal to \(1\): \(x^2-x-1=1\), so \(x^2-x-2=0\). 4. Factor: \((x-2)(x+1)=0\). Thus, \(x=2\) or \(x=-1\). 5. Evaluate the function: \(f(2)=-\frac{4}{3}\) and \(f(-1)=\frac{1}{6}\). The points are \((2,-\frac{4}{3})\) and \((-1,\frac{1}{6})\).

\((2,-\frac{4}{3})\) and \((-1,\frac{1}{6})\)

- Set the two function values equal to find the intersections. - Differentiate both polynomials term by term to obtain their tangent slopes. - Use the angle-between-lines formula for the two slopes at each intersection. - Round only the final angle to the requested precision.

1. Set the functions equal: \(x^2-2x+2=-x^2+4x-2\). This simplifies to \(x^2-3x+2=0\), so \(x=1\) or \(x=2\). The points are \((1,1)\) and \((2,2)\). 2. Differentiate term by term: \(f'(x)=2x-2\) and \(g'(x)=-2x+4\). 3. At \(x=1\), the slopes are \(0\) and \(2\). Thus, \(\tan(\gamma)=2\), so \(\gamma=\arctan(2)\approx63.43^\circ\). 4. At \(x=2\), the slopes are \(2\) and \(0\), giving the same acute angle, \(63.43^\circ\).

1. \((1,1)\) and \((2,2)\) 2. The acute intersection angle is approximately \(63.43^\circ\) at both points.

- Differentiate both functions term by term and evaluate the slopes at each intersection. - Equal tangent slopes give an intersection angle of \(0^\circ\). - Otherwise use the acute angle-between-lines formula and the requested precision.

1. Differentiate term by term: \(f'(x)=3x^2-1\) and \(g'(x)=2x\). 2. At \(P=(-1,0)\), the slopes are \(2\) and \(-2\). Therefore, \(\tan(\gamma)=\frac{4}{3}\), so \(\gamma=\arctan\!\left(\frac{4}{3}\right)\approx53.13^\circ\). 3. At \(Q=(1,0)\), both slopes equal \(2\). The tangent lines coincide, so the intersection angle is \(0^\circ\); the graphs are tangent at \(Q\).

At \(P=(-1,0)\), the acute intersection angle is approximately \(53.13^\circ\). At \(Q=(1,0)\), the angle is \(0^\circ\); the graphs are tangent there.

- Parallel lines have equal slopes. - Express each tangent slope using a derivative. - Set the derivative expressions equal. - Solve the resulting quadratic equation.

1. Differentiate: \(f'(x)=x^2-4\) and \(g'(x)=2x-1\). 2. Parallel tangent lines have equal slopes, so solve \(x^2-4=2x-1\). 3. Rearrange: \(x^2-2x-3=0\). 4. Factor: \((x-3)(x+1)=0\). Therefore, \(x=-1\) or \(x=3\).

\(x=-1\) and \(x=3\)

- Equal slopes mean equal derivative values. - Set the derivative functions equal and solve. - For each tangent line, use its own point on the corresponding graph. - Use point-slope form.

1. Differentiate: \(f'(x)=3x^2-6x\) and \(g'(x)=6x-9\). 2. Set the slopes equal: \(3x^2-6x=6x-9\). Dividing by \(3\) gives \(x^2-4x+3=0\), so \(x=1\) or \(x=3\). 3. The smaller value is \(x=1\). The common slope is \(-3\). 4. Since \(f(1)=3\), the tangent to \(f\) is \(y-3=-3(x-1)\), or \(y=-3x+6\). 5. Since \(g(1)=-4\), the tangent to \(g\) is \(y+4=-3(x-1)\), or \(y=-3x-1\).

a) \(x=1\) and \(x=3\) b) At \(x=1\), the tangent lines are \(y=-3x+6\) and \(y=-3x-1\).

- Set the function values equal to locate the common point. - Expand the translated square before differentiating so only the rules from this topic are needed. - Find each tangent line from its point and slope, then set \(y=0\) for the x-intercepts. - Use the x-axis segment as the triangle's base.

1. Set the functions equal: \(x^2=(x-2)^2\). This gives \(x=1\), and the common value is \(1\), so \(P=(1,1)\). 2. Expand \(g(x)=x^2-4x+4\). Then \(f'(x)=2x\) and \(g'(x)=2x-4\) by the rules available in this topic. At \(x=1\), the slopes are \(2\) and \(-2\). 3. The tangent lines are \(y=2x-1\) and \(y=-2x+3\). 4. Their x-intercepts are \(\left(\frac{1}{2},0\right)\) and \(\left(\frac{3}{2},0\right)\). 5. The base on the x-axis has length \(1\), and the height from \(P\) is \(1\), so the area is \(\frac{1}{2}\) square unit.

a) \(P=(1,1)\) b) \(\left(\frac{1}{2},0\right)\) and \(\left(\frac{3}{2},0\right)\) c) \(\frac{1}{2}\) square unit

- How can you tell when two graphs pass through the same point? - What equation expresses equal tangent slopes at a common input? - For the normal, what point and direction must its line equation use?

1. Set the functions equal: \(x^2+3x=-x^2+5\), so \(2x^2+3x-5=0\). The solutions are \(x=1\) and \(x=-\frac{5}{2}\). 2. The intersection points are \((1,4)\) and \(\left(-\frac{5}{2},-\frac{5}{4}\right)\). 3. Differentiate: \(f'(x)=2x+3\) and \(g'(x)=-2x\). 4. Equal tangent slopes require \(2x+3=-2x\), giving \(x=-\frac{3}{4}\). 5. At \(x=2\), \(g(2)=1\) and the tangent slope is \(-4\). The normal slope is \(\frac{1}{4}\). 6. The normal line is \(y-1=\frac{1}{4}(x-2)\), or \(y=\frac{1}{4}x+\frac{1}{2}\).

a) \((1,4)\) and \(\left(-\frac{5}{2},-\frac{5}{4}\right)\) b) \(x=-\frac{3}{4}\) c) \(y=\frac{1}{4}x+\frac{1}{2}\)

- A tangent parallel to the x-axis has slope \(0\). - Write the tangent line at \(x=a\) using point-slope form. - Treat \(a\) as a fixed parameter while simplifying. - Set \(x=0\) to find the y-intercept.

1. Differentiate: \(h'(x)=\frac{1}{2}x-1\). A horizontal tangent has slope \(0\), so \(\frac{1}{2}x-1=0\), giving \(x=2\). 2. Since \(h(2)=-1\), the point is \(P=(2,-1)\). 3. At \(x=a\), the slope is \(\frac{1}{2}a-1\), and the point is \((a,\frac{1}{4}a^2-a)\). 4. Point-slope form gives \(y=(\frac{1}{2}a-1)(x-a)+\frac{1}{4}a^2-a\). 5. Simplifying gives \(t_a: y=(\frac{1}{2}a-1)x-\frac{1}{4}a^2\). 6. Setting \(x=0\) gives \(y=-\frac{1}{4}a^2\), so the y-intercept is \((0,-\frac{1}{4}a^2)\).

a) \(P=(2,-1)\) b) \(t_a: y=(\frac{1}{2}a-1)x-\frac{1}{4}a^2\) c) Setting \(x=0\) gives \(S=(0,-\frac{1}{4}a^2)\).

- The tangent slope at \(x=-1\) is \(g'(-1)\). - Use the given slope to form an equation containing only \(a\). - Find the point of tangency after determining \(a\). - Use the point-slope form of a line.

1. Differentiate: \(g'(x)=2ax+4\). 2. Use the given slope: \(g'(-1)=2\). Therefore, \(-2a+4=2\), so \(a=1\). 3. With \(a=1\), the point of tangency is \((-1,g(-1))=(-1,-5)\). 4. A line with slope \(2\) through \((-1,-5)\) satisfies \(y+5=2(x+1)\), so \(y=2x-3\).

a) \(a=1\) b) \(y=2x-3\)

- Set the derivative equal to the required slope. - Solve for every possible input. - Substitute each input into the original function. - Use the common slope and each point to write the tangent lines.

1. Differentiate: \(f'(x)=3x^2-6x-6\). 2. Set the derivative equal to \(3\): \(3x^2-6x-6=3\). Dividing by \(3\) gives \(x^2-2x-3=0\). 3. Factor: \((x-3)(x+1)=0\), so \(x=3\) or \(x=-1\). 4. Evaluate the function: \(f(3)=-16\) and \(f(-1)=4\). The points are \((3,-16)\) and \((-1,4)\). 5. At \((3,-16)\), the tangent is \(y+16=3(x-3)\), or \(y=3x-25\). 6. At \((-1,4)\), the tangent is \(y-4=3(x+1)\), or \(y=3x+7\).

The points are \((3,-16)\) and \((-1,4)\). The tangent lines are \(y=3x-25\) and \(y=3x+7\), respectively.

- What must a function value equal where its graph crosses the x-axis? - Recall the power rule, including its use with negative exponents. - What relationship between two nonzero slopes indicates perpendicular tangent lines? - Compare the products of the derivatives at the common point.

1. Solve \(0.5x^2-2=0\): \(x^2=4\), so \(x=\pm2\). The positive zero is \(x_0=2\). 2. Evaluate the other functions at \(x=2\): \(g(2)=-0.125\cdot2^2+0.5=0\) and \(h(2)=\frac{2}{2}-1=0\). Therefore, all three graphs pass through \((2,0)\). 3. Differentiate: \(f'(x)=x\), \(g'(x)=-0.25x\), and \(h'(x)=-\frac{2}{x^2}\). At \(x=2\), the slopes are \(f'(2)=2\) and \(g'(2)=h'(2)=-0.5\). Since \(2\cdot(-0.5)=-1\), each of the other two graphs intersects the graph of \(f\) at a right angle.

1. \(x_0=2\) 2. Yes. All three graphs cross the x-axis at \((2,0)\). 3. Yes. The products of the tangent slopes are \(f'(2)g'(2)=-1\) and \(f'(2)h'(2)=-1\).

- Use only odd powers for origin symmetry. - Translate the slope condition into an equation involving the derivative. - Leave one coefficient as a parameter when finding all possible functions. - Use the additional point to determine the parameter.

1. A cubic polynomial symmetric about the origin has the form \(f(x)=ax^3+cx\), where \(a\ne0\). 2. Since \(f'(x)=3ax^2+c\), the slope condition \(f'(1)=6\) gives \(3a+c=6\), so \(c=6-3a\). 3. Therefore, all such cubic polynomials are \(f_a(x)=ax^3+(6-3a)x\), where \(a\in\mathbb{R}\setminus\{0\}\). 4. To pass through \(Q(2, 20)\), the polynomial must satisfy \(8a+2(6-3a)=20\). Thus, \(2a+12=20\), so \(a=4\) and \(c=-6\). 5. The required polynomial is \(f(x)=4x^3-6x\).

All such polynomials: \(f_a(x)=ax^3+(6-3a)x\), where \(a\in\mathbb{R}\setminus\{0\}\) Polynomial through \(Q(2, 20)\): \(f(x)=4x^3-6x\)

- A common tangent requires a common point and a common derivative value. - First find the input that makes the parameter-dependent term vanish. - Expand the shifted cube before differentiating so only previously established rules are needed.

1. For the function value to be independent of \(k\), the parameter term must be zero. Since \((x-2)^3=0\), use \(x=2\). 2. Evaluate the function: \(g_k(2)=5\). Every graph passes through \((2,5)\). 3. Expand the parameter term before differentiating: \(g_k(x)=k(x^3-6x^2+12x-8)+x^2+1\). 4. Apply the power, sum, and constant-multiple rules: \(g_k'(x)=k(3x^2-12x+12)+2x\). 5. At \(x=2\), the coefficient of \(k\) is \(12-24+12=0\), so \(g_k'(2)=4\), independent of \(k\). 6. The line with slope \(4\) through \((2,5)\) is \(y-5=4(x-2)\), or \(y=4x-3\).

Every graph has the same tangent line at \((2,5)\), and the line is \(y=4x-3\).

- Tangency to the x-axis gives both a zero and a zero slope. - Use the origin to determine the constant term. - Translate each remaining condition into an equation. - Solve the resulting linear system.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Passing through the origin gives \(d=0\). 2. The point \(P(2, 4)\) gives \(8a+4b+2c=4\). Tangency to the x-axis at \(x=4\) gives \(f(4)=0\) and \(f'(4)=0\), so \(64a+16b+4c=0\) and \(48a+8b+c=0\). 3. Solving the system gives \(a=\frac{1}{2}\), \(b=-4\), and \(c=8\). 4. Therefore, \(f(x)=\frac{1}{2}x^3-4x^2+8x\).

\(f(x)=\frac{1}{2}x^3-4x^2+8x\)

- A tangent line through the origin has equation \(y=mx\). - At the point of tangency, use \(m=f'(x_0)\) and \(f(x_0)=mx_0\). - Perpendicular nonvertical lines have slopes whose product is \(-1\).

1. Differentiate: \(f'(x)=\frac{3}{2}x^2-4x+1\). 2. A tangent line through the origin at \(x=x_0\) must satisfy \(f(x_0)=f'(x_0)x_0\). 3. Substitute: \(\frac{1}{2}x_0^3-2x_0^2+x_0=x_0\left(\frac{3}{2}x_0^2-4x_0+1\right)\). 4. Simplifying gives \(x_0^2(x_0-2)=0\). The solutions are \(x_0=0\) and \(x_0=2\). Since the point of tangency is not the origin, \(x_0=2\). 5. Since \(f(2)=-2\), the point of tangency is \(B(2, -2)\). The tangent slope is \(f'(2)=-1\), so the tangent line is \(t: y=-x\). 6. At the origin, the tangent slope is \(f'(0)=1\). Because \((-1)(1)=-1\), the two tangent lines are perpendicular.

a) \(B(2, -2)\); \(t: y=-x\) b) The slopes are \(-1\) and \(1\), whose product is \(-1\), so the tangent lines are perpendicular.

- Set the two function expressions equal to locate the intersections. - Differentiate both formulas term by term to obtain tangent slopes. - Use the acute-angle formula only after the two slopes at each intersection are known.

1. Set the functions equal: \(x^2-6x+9=-x+5\). Then \(x^2-5x+4=0\), so \(x=1\) or \(x=4\). Thus, \(P_1=(1,4)\) and \(P_2=(4,1)\). 2. Differentiate term by term: \(f'(x)=2x-6\) and \(g'(x)=-1\). At \(P_1\), the slopes are \(-4\) and \(-1\). At \(P_2\), the slopes are \(2\) and \(-1\). 3. Using \(\tan(\alpha)=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\), at \(P_1\), \(\tan(\alpha_1)=0.6\), so \(\alpha_1\approx30.96^\circ\). 4. At \(P_2\), \(\tan(\alpha_2)=3\), so \(\alpha_2\approx71.57^\circ\).

a) \(P_1=(1,4)\) and \(P_2=(4,1)\) b) At \(P_1\), the slopes are \(-4\) for \(f\) and \(-1\) for \(g\). At \(P_2\), the slopes are \(2\) for \(f\) and \(-1\) for \(g\). c) \(\alpha_1\approx30.96^\circ\) and \(\alpha_2\approx71.57^\circ\)

- Write the tangent-line equation at a general input \(u\). - A tangent through \(Q\) must produce the point's y-value when \(x=0\). - Solve the resulting equation for the possible tangency inputs. - Use each input in the original function and derivative to finish the tangent line.

1. The tangent line at \(x=u\) is \(t(x)=f'(u)(x-u)+f(u)\), with \(f'(x)=-0.5x+1\). 2. Requiring \(t(0)=7\) gives \((-0.5u+1)(-u)-0.25u^2+u+3=7\), which simplifies to \(0.25u^2=4\). Thus, \(u=4\) or \(u=-4\). 3. Since \(f(4)=3\) and \(f(-4)=-5\), the tangency points are \(B_1=(4,3)\) and \(B_2=(-4,-5)\). 4. The corresponding slopes are \(-1\) and \(3\), so the tangent lines are \(y=-x+7\) and \(y=3x+7\).

a) \(B_1=(4,3)\) and \(B_2=(-4,-5)\) b) \(t_1:y=-x+7\) and \(t_2:y=3x+7\)

- The derivative gives the slope of the graph at an input. - Translate each requested tangent condition into an equation involving \(f'(x)\). - After solving for an input, use the original function to obtain the corresponding point.

1. Differentiate: \(f'(x)=\frac{1}{2}x^2-2x\). 2. Set the derivative equal to \(-2\): \(\frac{1}{2}a^2-2a=-2\). This gives \(a^2-4a+4=0\), so \(a=2\). 3. Evaluate the function: \(f(2)=-\frac{2}{3}\). Thus, \(A\left(2,-\frac{2}{3}\right)\). 4. A line parallel to \(y=6x-1\) has slope \(6\). Set \(\frac{1}{2}x^2-2x=6\). Then \(x^2-4x-12=0\), so \(x=6\) or \(x=-2\). 5. Evaluate the function: \(f(6)=2\) and \(f(-2)=-\frac{10}{3}\). The points are \((6,2)\) and \(\left(-2,-\frac{10}{3}\right)\).

a) \(f'(x)=\frac{1}{2}x^2-2x\) b) \(A\left(2,-\frac{2}{3}\right)\) c) \((6,2)\) and \(\left(-2,-\frac{10}{3}\right)\)

- Read two exact points from the graph and substitute them into \(f(x)=ax^2+bx\). - Evaluate \(f(3)\) and use the derivative for the tangent slope. - Convert the tangent slope to the normal slope. - Use the normal line's axis intercepts as the triangle's legs.

1. From the graph, use \((2,2)\) and \((4,0)\). Substitution gives \(4a+2b=2\) and \(16a+4b=0\). 2. Simplifying gives \(2a+b=1\) and \(4a+b=0\). Subtracting yields \(a=-0.5\), and then \(b=2\). Thus, \(f(x)=-0.5x^2+2x\). 3. Evaluate the function: \(f(3)=1.5\), so \(P=(3,1.5)\). Differentiate: \(f'(x)=-x+2\), so the tangent slope at \(x=3\) is \(-1\). 4. The normal slope is \(1\). Through \(P\), the normal is \(y-1.5=x-3\), or \(y=x-1.5\). 5. Its intercepts are \((0,-1.5)\) and \((1.5,0)\). The triangle area is \(\frac{1}{2}(1.5)(1.5)=1.125=\frac{9}{8}\) square units.

a) \(a=-0.5\), \(b=2\) b) \(n: y=x-1.5\) c) \(\frac{9}{8}\) square units

- Compute both derivatives before matching any panels. - Use the formulas you obtained to identify the line and parabola representing the derivatives. - Confirm each pairing by comparing derivative zeros with horizontal tangents of the original graph.

1. Differentiate term by term: \(f'(x)=x\) and \(g'(x)=-0.6x^2+1.5\). 2. Graph (3) is the upward-opening parabola with vertex \((0,-2)\), so it represents \(f\). Graph (1) is the line \(y=x\), so it represents \(f'\). 3. Graph (2) is the odd cubic with negative leading coefficient, so it represents \(g\). Graph (4) is the downward-opening parabola with y-intercept \(1.5\), so it represents \(g'\). 4. The zero of graph (1) at \(x=0\) matches the horizontal tangent of graph (3) there. The zeros of graph (4) align with the turning points of graph (2).

a) \(f'(x)=x\); \(g'(x)=-0.6x^2+1.5\) b) (1) \(f'\); (2) \(g\); (3) \(f\); (4) \(g'\) c) The derivative zeros align with horizontal tangents of their corresponding function graphs.

- Set the function equal to zero to find the x-intercepts. - Use the derivative for each tangent slope. - Solve the two tangent equations together to locate \(S\). - Use \(\overline{PQ}\) as the triangle's base.

1. Set \(f(x)=0\): \(-0.25x^2+4=0\), so \(x=\pm4\). Thus, \(P=(-4,0)\) and \(Q=(4,0)\). 2. Differentiate: \(f'(x)=-0.5x\). At \(P\), the slope is \(2\), so \(t_1:y=2x+8\). At \(Q\), the slope is \(-2\), so \(t_2:y=-2x+8\). 3. Solving the tangent equations together gives \(S=(0,8)\). 4. Triangle \(PQS\) has base \(8\) and height \(8\), so its area is \(32\) square units.

a) \(P=(-4,0)\) and \(Q=(4,0)\) b) \(t_1:y=2x+8\) and \(t_2:y=-2x+8\) c) \(32\) square units

- Differentiate \(f\) before trying to match the panels. - Use the computed derivative formula to identify the linear panel. - Then use vertical and horizontal translations to distinguish \(g\) and \(h\). - Check that a zero of the derivative aligns with a horizontal tangent of \(f\).

1. Differentiate term by term: \(f'(x)=2x-2\). 2. The graph of \(f\) is an upward-opening parabola with zeros at \(0\) and \(2\) and vertex \((1,-1)\), so it is graph II. 3. The derivative \(f'(x)=2x-2\) is a line with zero at \(x=1\), so it is graph IV. Its zero aligns with the horizontal tangent at the vertex of graph II. 4. The graph of \(g=f+3\) is shifted up \(3\) units and has vertex \((1,2)\), so it is graph I. 5. The graph of \(h(x)=f(x+2)\) is shifted left \(2\) units and has vertex \((-1,-1)\), so it is graph III.

a) \(f'(x)=2x-2\) b) Graph I: \(g\); Graph II: \(f\); Graph III: \(h\); Graph IV: \(f'\) c) Graph IV crosses the x-axis at \(x=1\), the x-coordinate of the horizontal tangent at the vertex of graph II.

- Read both y-values at \(x=4\). - Use the actual endpoints of each drawn tangent segment to compute rise over run. - Compare the slope ratio with the function-value scale factor.

1. From the graph, \(f(4)=2\) and \(g(4)=4\). Since \(g(4)=kf(4)\), \(4=2k\), so \(k=2\). 2. The tangent to \(f\) rises \(4\) units over a run of \(4\), so \(f'(4)=1\). The tangent to \(g\) rises \(6\) units over a run of \(3\), so \(g'(4)=2\). 3. Therefore, \(g'(4)=2f'(4)=kf'(4)\). This illustrates the constant multiple rule: if \(g(x)=kf(x)\), then \(g'(x)=kf'(x)\).

1. \(k=2\) 2. \(f'(4)=1\) and \(g'(4)=2\) 3. \(g'(4)=kf'(4)\), illustrating the constant multiple rule.

- Set the two function expressions equal to locate the intersections. - Differentiate both polynomials to obtain tangent slopes. - Use the acute angle-between-lines formula and round only the final angle.

1. Set the functions equal: \(x^2-1=7-x^2\), giving \(x=\pm2\). The intersection points are \((2,3)\) and \((-2,3)\). 2. Differentiate: \(f'(x)=2x\) and \(g'(x)=-2x\). At \(x=2\), the tangent slopes are \(4\) and \(-4\). 3. Thus, \(\tan(\phi)=\left|\frac{4-(-4)}{1+4(-4)}\right|=\frac{8}{15}\), so \(\phi\approx28.1^\circ\). 4. By symmetry, the acute angle at the other intersection is also \(28.1^\circ\).

a) \((2,3)\) and \((-2,3)\) b) The angle at each intersection is approximately \(28.1^\circ\).

- Set the two function expressions equal to find all intersections. - Differentiate both functions to obtain the tangent slopes at each common input. - Use the acute angle-between-lines relation once the slopes are known.

1. Set \(\frac{1}{4}x^3=x\). Then \(x(x^2-4)=0\), so the intersection points are \((-2,-2)\), \((0,0)\), and \((2,2)\). 2. The tangent slopes are given by \(f'(x)=\frac{3}{4}x^2\) and \(g'(x)=1\). 3. At \((0,0)\), the slopes are \(0\) and \(1\), so the acute angle is \(45^\circ\). 4. At \((-2,-2)\) and \((2,2)\), the slopes are \(3\) and \(1\). Thus, \(\tan\theta=\frac{1}{2}\), so \(\theta=\arctan\!\left(\frac{1}{2}\right)\approx26.6^\circ\).

a) \((-2,-2)\), \((0,0)\), and \((2,2)\) b) At \((0,0)\), the acute angle is \(45^\circ\). At the other two points, it is \(\arctan\!\left(\frac{1}{2}\right)\approx26.6^\circ\).

- Use the midpoint coordinates in the cable model. - The derivative gives the slope of the cable at the tower. - Relate slope to angle using the tangent function. - For an upward-opening parabola, the minimum occurs at the vertex.

1. At the midpoint, \(x=0\). Set the height equal to \(30\): \(50-2500k=30\). Thus, \(k=\frac{20}{2500}=\frac1{125}\). 2. The right tower is at \(x=50\). Differentiate: \(f_k'(x)=2kx\). For \(k=\frac1{125}\), the slope at the tower is \(f_k'(50)=\frac45\). 3. If \(\theta\) is the angle of inclination, then \(\tan(\theta)=\frac45\), so \(\theta=\arctan\left(\frac45\right)\approx38.66^\circ\). 4. Because \(k>0\), the lowest point is at \(x=0\). The cable touches or passes below the roadway when \(50-2500k\le0\). Therefore, \(k\ge\frac1{50}\). It first touches the roadway when \(k=\frac1{50}\).

a) \(k=\frac1{125}\) b) \(\theta\approx38.66^\circ\) c) \(k\ge\frac1{50}\); it first touches the roadway at \(k=\frac1{50}\).

- Differentiate the polynomial term by term. - A horizontal tangent has slope \(0\). - A tangent parallel to \(y=x\) must have slope \(1\). - Substitute each solved input into \(f\) to recover the point.

1. Differentiate term by term: \(f'(x)=-x+3\). 2. For a horizontal tangent, solve \(-x+3=0\), giving \(x=3\). Since \(f(3)=4.5\), \(S=(3,4.5)\). 3. The line \(y=x\) has slope \(1\). Solve \(-x+3=1\), giving \(x=2\). Since \(f(2)=4\), \(R=(2,4)\).

a) \(f'(x)=-x+3\) b) \(S=(3,4.5)\) c) \(x=2\), and \(R=(2,4)\)

- Differentiate \(f\) and \(g\) first. - Compare the degree and symmetry of each derivative with the graphs. - Find the zeros of each derivative and match them to the \(x\)-intercepts shown.

1. Differentiate each function using the power rule: \(f'(x) = x^2 - 1\) and \(g'(x) = x^3 - 4x\). 2. The graph of \(f'\) is an upward-opening parabola, symmetric about the \(y\)-axis, with vertex \((0, -1)\) and zeros at \(x=-1\) and \(x=1\). This is graph A. 3. Since \(g'(x)=x(x-2)(x+2)\), its graph is an odd cubic with zeros at \(x=-2\), \(x=0\), and \(x=2\). This is graph B.

Graph A represents \(f'\), and graph B represents \(g'\).

- For part a), what happens to the slope calculation when the two x-coordinates are equal? - A line through two points and a tangent at one of those points need not have the same slope. - In part b), think about what it means for a line to contain the origin.

1. Since \(Q_a=\left(a,\frac{1}{2}a^2+2\right)\), the slope of the line through \(P\) and \(Q_a\) is \(m_a=\frac{a^2+4}{2a}\). 2. Differentiate: \(f'(x)=x\). 3. The tangent at \(Q_a\) is \(y=a(x-a)+\frac{1}{2}a^2+2=ax-\frac{1}{2}a^2+2\). 4. For this line to pass through \(P=(0,0)\), \(0=-\frac{1}{2}a^2+2\). Thus, \(a=2\) or \(a=-2\).

a) \(m_a=\frac{a^2+4}{2a}\) b) \(a=2\) and \(a=-2\)

- Use the factored form of a quadratic with the two given zeros. - Differentiate that form after expanding it. - Write the tangent line through the origin using its derivative slope. - Use the given intersection point of the tangents to determine the remaining coefficient.

1. Since the zeros are \(0\) and \(4\), write \(f(x)=ax(x-4)=ax^2-4ax\). 2. Differentiate: \(f'(x)=2ax-4a\). At the origin, the tangent slope is \(-4a\), so the tangent line there is \(y=-4ax\). 3. Since \(S=(2,-6)\) lies on this tangent, \(-6=-8a\), so \(a=\frac{3}{4}\). 4. Therefore, \(f(x)=\frac{3}{4}x^2-3x\).

\(f(x)=\frac{3}{4}x^2-3x\)

- Interpret a function value as height and a derivative value as slope. - Use vertex form because the highest point is given. - The water reaches the surface at a zero of the function. - The tangent of the launch angle equals the slope at the nozzle.

1. The condition \(f(0)=2\) means the water leaves the nozzle \(2\,\text{ft}\) above the surface. The condition \(f'(4)=0\) means the path has a horizontal tangent at a horizontal distance of \(4\,\text{ft}\), where it reaches its maximum height. 2. Using vertex form with vertex \((4, 10)\), write \(f(x)=a(x-4)^2+10\). Since \(f(0)=2\), \(16a+10=2\), so \(a=-\frac{1}{2}\). Thus, \(f(x)=-\frac{1}{2}(x-4)^2+10=-\frac{1}{2}x^2+4x+2\). 3. Set \(f(x)=0\): \(-\frac{1}{2}x^2+4x+2=0\), or \(x^2-8x-4=0\). The positive solution is \(x=4+2\sqrt{5}\approx8.47\). The stream hits the water about \(8.47\,\text{ft}\) from the nozzle. 4. The initial slope is \(f'(0)\). Since \(f'(x)=-x+4\), \(f'(0)=4\). Therefore, \(\tan(\theta)=4\), so \(\theta=\arctan(4)\approx76.0^\circ\).

a) \(f(0)=2\) gives the nozzle's height, and \(f'(4)=0\) identifies the horizontal tangent at the maximum height. b) \(f(x)=-\frac{1}{2}x^2+4x+2\) c) \(4+2\sqrt{5}\,\text{ft}\approx8.47\,\text{ft}\) d) \(\theta\approx76.0^\circ\)

- Read the derivative of \(u\) from its tangent line at the marked input. - Because \(v\) is a line, its derivative is its constant slope. - Apply the difference rule after obtaining the two component slopes.

1. The tangent line \(t\) to \(u\) passes through grid points \((0.5,0)\) and \((1,1)\), so \(u'(1)=\frac{1-0}{1-0.5}=2\). 2. The graph of \(v\) is a line with slope \(-1\), so \(v'(1)=-1\). 3. By the difference rule, \(w'(1)=u'(1)-v'(1)=2-(-1)=3\).

\(w'(1)=3\)

- A tangent at input \(u\) has slope \(h'(u)\). - Write the tangent line at the unknown point of tangency. - Require that line to pass through \(P=(2,0)\). - The resulting equation may have more than one solution for \(u\).

1. Differentiate: \(h'(x)=2x-4\). 2. Let the point of tangency be \(B=(u,h(u))\). A tangent line at \(u\) is \(y=(2u-4)(x-u)+u^2-4u+5\). 3. Require the line to pass through \(P=(2,0)\): \(0=(2u-4)(2-u)+u^2-4u+5\). 4. Simplifying gives \(u^2-4u+3=0\), so \(u=1\) or \(u=3\). 5. The points of tangency are \((1,h(1))=(1,2)\) and \((3,h(3))=(3,2)\). 6. At \(u=1\), the slope is \(-2\), giving \(y=-2(x-2)=-2x+4\). At \(u=3\), the slope is \(2\), giving \(y=2(x-2)=2x-4\).

The points of tangency are \((1,2)\) and \((3,2)\). The tangent lines are \(y=-2x+4\) and \(y=2x-4\), respectively.

- First find the two x-intercepts of the parabola. - Use the derivative to write both tangent lines. - Their intersection is the third vertex of the triangle. - Use the x-axis segment as the base when computing the area.

1. Factor \(f_k(x)=x(x-k)\). The x-intercepts are \(O=(0,0)\) and \(A=(k,0)\). 2. Differentiate: \(f_k'(x)=2x-k\). The tangent slopes are \(f_k'(0)=-k\) and \(f_k'(k)=k\). 3. The tangent lines are \(y=-kx\) and \(y=k(x-k)=kx-k^2\). 4. Set the equations equal: \(-kx=kx-k^2\). Since \(k>0\), \(x=\frac{k}{2}\). Then \(y=-\frac{k^2}{2}\), so \(S=(\frac{k}{2},-\frac{k^2}{2})\). 5. The base \(OA\) has length \(k\), and the height is \(\frac{k^2}{2}\). Thus, the area is \(\frac{1}{2}\cdot k\cdot\frac{k^2}{2}=\frac{k^3}{4}\). 6. Solve \(\frac{k^3}{4}=16\). This gives \(k^3=64\), so \(k=4\).

\(k=4\)

- Treat \(k\) as a constant when differentiating and simplifying. - Use the point-slope form of a tangent line at \(x=x_0\). - For the triangle, find the length of the side on the x-axis and the vertical distance from the third vertex to the x-axis. - Lengths and areas must be positive.

1. Solve \(x^3 - kx = 0\): \(x(x^2-k)=0\), so the zeros are \(x=0\) and \(x=\pm\sqrt{k}\). Therefore, \(x_P=\sqrt{k}\). 2. Differentiate: \(f_k'(x)=3x^2-k\). 3. At the origin, \(f_k'(0)=-k\). Since the tangent passes through \((0, 0)\), \(t_0: y=-kx\). 4. At \(P(\sqrt{k}, 0)\), the slope is \(f_k'(\sqrt{k})=3k-k=2k\). Thus \(t_P: y=2k(x-\sqrt{k})\). 5. Find the intersection of the tangent lines by solving \(-kx=2k(x-\sqrt{k})\). Since \(k>0\), \(x=\frac{2}{3}\sqrt{k}\), and then \(y=-\frac{2}{3}k\sqrt{k}\). 6. The triangle has base \(\sqrt{k}\) and height \(\frac{2}{3}k\sqrt{k}\). Therefore, \(A(k)=\frac{1}{2}\cdot\sqrt{k}\cdot\frac{2}{3}k\sqrt{k}=\frac{1}{3}k^2\).

a) \(x_P=\sqrt{k}\) b) \(t_0: y=-kx\); \(t_P: y=2k(x-\sqrt{k})\) c) The tangent lines intersect at \(\left(\frac{2}{3}\sqrt{k}, -\frac{2}{3}k\sqrt{k}\right)\), giving \(A(k)=\frac{1}{3}k^2\).

- Use the fact that \(P\) lies on the graph. - What can the triangle's area tell you about the tangent line? - How does the tangent's slope relate to the derivative of the curve at \(P\)?

1. Since \(P(2,0)\) lies on the graph, \(f_{a,b}(2)=0\). Therefore, \(4a+2b+5=0\), or \(4a+2b=-5\). 2. The tangent line crosses the x-axis at \((2,0)\). Let its y-intercept be \((0,y_0)\). The triangle's area is \(\frac{1}{2}\cdot2\cdot y_0=6\), so \(y_0=6\). 3. The tangent line through \((0,6)\) and \((2,0)\) has slope \(\frac{0-6}{2-0}=-3\). 4. Differentiate: \(f_{a,b}'(x)=2ax+b\). At \(x=2\), the tangent slope gives \(4a+b=-3\). 5. Solve the system \(4a+2b=-5\) and \(4a+b=-3\). Subtraction gives \(b=-2\), and then \(a=-\frac{1}{4}\).

\(a=-\frac{1}{4}\) and \(b=-2\)

- Differentiate the polynomial and write the tangent equation at a general input \(a\). - Substitute the coordinates of \(P\) into that general line. - Factor the resulting equation in \(a\).

1. Differentiate term by term: \(f'(x)=1.5x^2-1.5\). 2. The tangent at \(x=a\) is \(t_a(x)=f'(a)(x-a)+f(a)\), which simplifies to \(t_a(x)=(1.5a^2-1.5)x-a^3\). 3. Requiring the tangent to pass through \(P=(2,-3)\) gives \(-3=2(1.5a^2-1.5)-a^3\), or \(a^2(3-a)=0\). 4. Thus, \(a=0\) or \(a=3\). The tangent lines are \(t_0(x)=-1.5x\) and \(t_3(x)=12x-27\).

a) \(t_a(x)=(1.5a^2-1.5)x-a^3\) b) \(a=0\) gives \(t_0(x)=-1.5x\), and \(a=3\) gives \(t_3(x)=12x-27\).

- A point on the path gives a condition on the function value. - At the highest point, the tangent is horizontal. - Start with a general quadratic and solve for its coefficients. - The tangent slope and the angle with the horizontal are related by tangent.

1. The takeoff condition is \(g(0)=0\). A horizontal tangent at the highest point gives \(g'(6)=0\). The remaining given point gives \(g(10)=5\). 2. Let \(g(x)=ax^2+bx+c\). From \(g(0)=0\), \(c=0\). Since \(g'(x)=2ax+b\), the condition \(g'(6)=0\) gives \(12a+b=0\), so \(b=-12a\). Using \(g(10)=5\), \(100a+10b=5\). Substitution gives \(-20a=5\), so \(a=-\frac{1}{4}\) and \(b=3\). Therefore, \(g(x)=-\frac{1}{4}x^2+3x\). 3. The maximum occurs at \(x=6\): \(g(6)=-\frac{1}{4}(36)+18=9\). The maximum height is \(9\,\text{m}\). 4. The slope at landing is \(g'(12)=-\frac{1}{2}(12)+3=-3\). If \(\theta\) is the magnitude of the angle with the horizontal ground, then \(\tan(\theta)=3\). Thus \(\theta=\arctan(3)\approx71.6^\circ\).

a) \(g(0)=0\) and \(g'(6)=0\). b) \(g(x)=-\frac{1}{4}x^2+3x\). c) \(9\,\text{m}\). d) \(\arctan(3)\approx71.6^\circ\).