Free math worksheets

- 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 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 \(f'(x)=20x^3\).

- 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 \(f'(0)=0\).

- Rewrite reciprocals and square roots as powers. - Differentiate each term separately. - Treat \(k\) as a constant. - Check where the original expressions and their derivatives are defined.

1. Differentiate each term in part a: \(\frac{d}{dx}(6x^{-1})=-6x^{-2}\), \(\frac{d}{dx}(4x^5)=20x^4\), and \(\frac{d}{dx}(-3\sin x)=-3\cos x\). Thus, \(f'(x)=-\frac{6}{x^2}+20x^4-3\cos x\), for \(x\ne0\). 2. For part b, \(\frac{d}{dx}(10x^{1/2})=5x^{-1/2}=\frac{5}{\sqrt{x}}\), \(\frac{d}{dx}\left(-\frac{1}{2}\cos x\right)=\frac{1}{2}\sin x\), and the derivative of \(12\) is \(0\). Therefore, \(f'(x)=\frac{5}{\sqrt{x}}+\frac{1}{2}\sin x\), for \(x>0\). 3. In part c, treat \(k\) as a constant: \(f'(x)=k\cos x-\frac{2}{x^2}\), for \(x\ne0\).

a) \(f'(x)=-\frac{6}{x^2}+20x^4-3\cos x\), for \(x\ne0\) b) \(f'(x)=\frac{5}{\sqrt{x}}+\frac{1}{2}\sin x\), for \(x>0\) c) \(f'(x)=k\cos x-\frac{2}{x^2}\), for \(x\ne0\)

- What is the derivative of a constant? - Substitute the two factors into the product rule. - What happens to the first product-rule term when one factor is constant? - Which method requires fewer steps?

1. By the constant multiple rule, \(f'(x)=8\cos(x)\). 2. Using the product rule, \(u'(x)=0\) and \(v'(x)=\cos(x)\), so \(f'(x)=u'(x)v(x)+u(x)v'(x)\) \(=0\sin(x)+8\cos(x)=8\cos(x)\). 3. The methods agree because the derivative of the constant factor is zero. The constant multiple rule is more efficient because it omits the product-rule term that equals zero.

1. \(f'(x)=8\cos(x)\) 2. \(f'(x)=8\cos(x)\). The product rule gives the same result because the derivative of \(8\) is \(0\); the constant multiple rule is more efficient because it avoids writing the zero product-rule term.

- 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 \(a=3\).

\(a=3\)

- 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) \(Q=(-1,-7)\)

- 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 \((-2,\frac{22}{3})\)

- 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. \(y=3x\)

- 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: \(\alpha\approx75.96^\circ\)

- Simplify the rational expression before differentiating. - Use the derivative value at \(x=1\) as the tangent slope. - For the asymptote, identify where the derivative denominator is zero while the numerator is nonzero. - For the sign question, analyze the numerator and denominator only near \(x=0\).

1. Simplify before differentiating: \(f(x)=4.5x-6+\frac{2}{x}\). 2. Differentiate term by term: \(f'(x)=4.5-\frac{2}{x^2}=\frac{9x^2-4}{2x^2}\). 3. At \(x=1\), the slope is \(f'(1)=2.5\). Through \((1,0.5)\), the tangent is \(y-0.5=2.5(x-1)\), so \(y=2.5x-2\). 4. The derivative is undefined at \(x=0\), and its numerator approaches \(-4\ne0\), so \(x=0\) is a vertical asymptote of \(f'\). For \(x\) sufficiently close to \(0\), \(9x^2-4<0\) while \(2x^2>0\), so \(f'(x)<0\) on both sides. Thus, there is no sign change across the asymptote sufficiently close to \(0\).

a) \(y=2.5x-2\) b) The vertical asymptote of \(f'\) is \(x=0\). The derivative is negative on both sides sufficiently close to the asymptote, so it does not change sign there.

- 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.

- 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

- 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 \(t_2:y=3x+7\)

- 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 \(28.1^\circ\).

- 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 \(R=(2,4)\)

- Use the slope formula between \(P\) and the parameterized point \(Q_a\). - The derivative at \(a\) gives the tangent slope. - Write the tangent line in terms of \(a\), then impose the condition that it passes through \(P\).

1. Since \(Q_a=\left(a,\frac{1}{2}a^2+2\right)\), the secant slope 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, \(f(x)=\frac{3}{4}x^2-3x\).

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

- 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, \(w'(1)=u'(1)-v'(1)=2-(-1)=3\).

\(w'(1)=3\)

- 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 \(t_3(x)=12x-27\).