- Think about how the tangent slopes of the sine and cosine curves change as \(x\) moves through \(0\). - Check your proposed rules against the slopes of both curves at \(x=0\).

1. The derivative of \(\sin x\) is \(\cos x\). 2. The derivative of \(\cos x\) is \(-\sin x\).

a) \(f'(x)=\cos x\) b) \(g'(x)=-\sin x\)

- Use the basic derivative formula for the natural exponential function. - Use the basic derivative formula for the natural logarithm on its positive domain. - Evaluate the derivative formulas at the requested inputs.

1. Since \(f'(x)=e^x\), \(f'(0)=e^0=1\). 2. Since \(g'(x)=\frac{1}{x}\), \(g'(1)=1\).

a) \(f'(0)=1\) b) \(g'(1)=1\)

- Recall how the derivative of \(a^x\) differs from the special case \(e^x\). - Recall the derivative formula for a logarithm with a base other than \(e\). - Keep the base-dependent constant factor in each derivative.

1. For a general exponential base, \(\frac{d}{dx}(a^x)=a^x\ln a\), so \(\frac{d}{dx}(7^x)=7^x\ln 7\). 2. For a general logarithm base, \(\frac{d}{dx}(\log_a x)=\frac{1}{x\ln a}\), so \(\frac{dy}{dx}=\frac{1}{x\ln 5}\) for \(x>0\).

a) \(7^x\ln 7\) b) \(\frac{dy}{dx}=\frac{1}{x\ln 5}\), for \(x>0\)

- What derivative rule applies to each trigonometric function? - What are the sine and cosine values at the given angle?

1. The derivative rules give \(f'(x)=\cos x\) and \(g'(x)=-\sin x\). 2. At \(x=\frac{\pi}{2}\), \(f'(\frac{\pi}{2})=0\) and \(g'(\frac{\pi}{2})=-1\). The sine tangent is horizontal; the cosine tangent falls.

\(f'(x)=\cos x\), so the sine tangent slope is \(0\), horizontal. \(g'(x)=-\sin x\), so the cosine tangent slope is \(-1\), falling.

- The base matters when differentiating an exponential function. What rule applies when the base is \(e\)? - Once you have a derivative, compare its value with \(f(0)\) and its formula with \(f(x)\).

1. The derivative rule gives \(f'(x)=e^x\). 2. The derivative equals the original function at every input.

\(f'(x)=e^x=f(x)\)

- Unlike a power of \(x\), the slope of \(\ln x\) becomes smaller as \(x\) grows. Does your derivative reflect that? - Check whether your result has a value at every input for which the original logarithm is defined.

1. The natural-logarithm rule gives \(f'(x)=\frac{1}{x}\) for \(x>0\).

\(f'(x)=\frac{1}{x}\), \(x>0\)

- The derivative rule for \(a^x\) includes a factor that depends on its base. - Compare your proposed rule for base \(5\) with the special case of base \(e\).

1. For a positive base \(a\), the rule is \(\frac{d}{dx}a^x=a^x\ln a\). 2. With \(a=5\), \(f'(x)=5^x\ln 5\).

\(f'(x)=5^x\ln 5\)

- A base-10 logarithm has a constant in its derivative that the natural logarithm does not. - Compare this base with the base used by the natural logarithm.

1. The logarithm rule for base \(a\) is \(\frac{d}{dx}\log_a x=\frac{1}{x\ln a}\). 2. With \(a=10\), \(f'(x)=\frac{1}{x\ln 10}\).

\(f'(x)=\frac{1}{x\ln 10}\), \(x>0\)

- Differentiate each term separately. - Recall the derivative of cosine. - Use the exact value of \(\sin\pi\). - Keep the final result exact.

1. Differentiate each term: \(f'(x)=4x+4\sin x+\frac{2}{\pi}\). 2. Evaluate at \(x=\pi\): \(f'(\pi)=4\pi+4\sin\pi+\frac{2}{\pi}\). Since \(\sin\pi=0\), \(f'(\pi)=4\pi+\frac{2}{\pi}\).

\(f'(x)=4x+4\sin x+\frac{2}{\pi}\); \(f'(\pi)=4\pi+\frac{2}{\pi}\)

- Differentiate sine and cosine carefully. - Track the sign when differentiating \(-2\cos x\). - Use unit-circle values at \(\frac{\pi}{2}\). - Simplify the polynomial term after substitution.

1. Differentiate each term: \(f'(x)=\frac{1}{2}\cos x+2\sin x+\frac{2}{\pi}x\). 2. Evaluate at \(x=\frac{\pi}{2}\): \(f'\left(\frac{\pi}{2}\right)=\frac{1}{2}\cos\left(\frac{\pi}{2}\right)+2\sin\left(\frac{\pi}{2}\right)+\frac{2}{\pi}\cdot\frac{\pi}{2}\). Using \(\cos\left(\frac{\pi}{2}\right)=0\) and \(\sin\left(\frac{\pi}{2}\right)=1\), \(f'\left(\frac{\pi}{2}\right)=3\).

\(f'(x)=\frac{1}{2}\cos x+2\sin x+\frac{2}{\pi}x\); \(f'\left(\frac{\pi}{2}\right)=3\)

- Differentiate each term separately. - Recall the derivatives of sine and cosine. - Treat numerical expressions such as \(\sin\pi\) as constants. - A constant denominator can be treated as a constant factor.

1. Apply the sum, difference, and constant multiple rules. 2. For part a, \(f'(x)=\frac{3\cos x+\sin x}{4}\). 3. For part b, \(f'(x)=10x+3\pi\cos x\). 4. For part c, \(f'(x)=2(3x^2+\sin x)=6x^2+2\sin x\). 5. Since \(\sin\pi=0\), part d simplifies to \(f(x)=-\cos x\). Therefore, \(f'(x)=\sin x\).

a) \(f'(x)=\frac{3\cos x+\sin x}{4}\) b) \(f'(x)=10x+3\pi\cos x\) c) \(f'(x)=6x^2+2\sin x\) d) \(f'(x)=\sin x\)

- Track the sign when differentiating cosine. - Constant terms disappear under differentiation. - Keep constant factors. - Evaluate \(\cos\left(\frac{\pi}{3}\right)\) before differentiating.

1. For part a, \(f'(x)=x^3+\sqrt{2}\sin x\). 2. For part b, \(f'(x)=3(\cos x-4)=3\cos x-12\). 3. Rewrite part c as \(\frac{1}{3}\sin x+4\). Then \(f'(x)=\frac{1}{3}\cos x\). 4. Since \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\), part d is \(\frac{1}{2}x^2+\sin x\). Therefore, \(f'(x)=x+\cos x\).

a) \(f'(x)=x^3+\sqrt{2}\sin x\) b) \(f'(x)=3\cos x-12\) c) \(f'(x)=\frac{1}{3}\cos x\) d) \(f'(x)=x+\cos x\)

- Differentiate every function before using the answer choices. - Treat the polynomial and trigonometric terms separately. - Check the sign change when differentiating cosine and keep constant factors attached to trig terms.

1. Differentiate \(f\) term by term: \(f'(x)=6x^2-\cos x=C(x)\). 2. Differentiate \(g\) term by term: \(g'(x)=12x^3-\sin x=B(x)\). 3. Differentiate \(h\) term by term: \(h'(x)=4-2\cos x=A(x)\).

\(f\to C\), \(g\to B\), \(h\to A\)

- The derivative gives the tangent slope. - Parallel lines have equal slopes. - Set the derivative equal to \(1.5\). - Use the unit circle to solve the cosine equation.

1. Differentiate: \(f'(x)=3\cos x\). Thus, \(f'\left(\frac{\pi}{3}\right)=3\cdot\frac{1}{2}=1.5\). 2. A parallel tangent must have slope \(1.5\), so \(3\cos x=1.5\). Therefore, \(\cos x=\frac{1}{2}\). In \([0, 2\pi]\), the solutions are \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\).

1. The slope is \(1.5\). 2. \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\)

- Determine the sign of the expression inside the logarithm before simplifying it. - Could rewriting the logarithm make its derivative easier to find? - Check that any logarithm identity you use holds throughout the domain you found.

1. The logarithm requires \(x^5e^3>0\). Since \(e^3>0\), this means \(x>0\), so the domain is \((0,\infty)\). 2. Use the product and power properties: \(f(x)=\ln(x^5)+\ln(e^3)=5\ln x+3\). 3. Differentiate: \(f'(x)=\frac{5}{x}\).

a) \((0,\infty)\) b) \(f(x)=5\ln x+3\) c) \(f'(x)=\frac{5}{x}\)

- Combine the square-root and logarithm domain restrictions. - Rewrite the square root as a fractional power. - Use the quotient property of logarithms. - How does the natural logarithm undo exponentiation with base \(e\)?

1. The square root requires \(x\ge0\), and the logarithm's argument must be positive. Therefore, the domain is \((0,\infty)\). 2. By logarithm properties, \(g(x)=\ln(\sqrt{x})-\ln(e^2)=\frac{1}{2}\ln x-2\). 3. Differentiate: \(g'(x)=\frac{1}{2x}\).

1. \((0,\infty)\) 2. \(g(x)=\frac{1}{2}\ln x-2\) 3. \(g'(x)=\frac{1}{2x}\)

- Split the logarithm of a product into a sum. - Simplify \(\ln(e^{3x})\). - Differentiate the logarithmic and linear terms separately.

1. Use logarithm properties: \(f(x)=\ln(x^2)+\ln(e^{3x})=2\ln x+3x\). 2. Differentiate: \(f'(x)=\frac{2}{x}+3\).

\(f'(x)=\frac{2}{x}+3\)

- Combine the restrictions from the root, denominator, and logarithm. - Use the quotient property of logarithms. - Rewrite the fourth root as a fractional power. - What happens when the natural logarithm is applied to a power of \(e\)?

1. The fourth root is in the denominator, and the logarithm's argument must be positive, so \(x>0\). The domain is \((0,\infty)\). 2. Simplify: \(g(x)=\ln(e^2)-\ln(x^{\frac{1}{4}})=2-\frac{1}{4}\ln x\). Differentiate: \(g'(x)=-\frac{1}{4x}\).

1. \((0,\infty)\) 2. \(g(x)=2-\frac{1}{4}\ln x\); \(g'(x)=-\frac{1}{4x}\)

- A constant added outside a function creates a vertical shift. - Differentiate the logarithm and the added constant separately. - For the tangent line, combine the function value and derivative value at \(x=1\).

1. Adding \(3\) shifts the graph of \(y=\ln x\) upward by \(3\) units. 2. The derivative of \(\ln x\) is \(\frac1x\), and the constant has derivative \(0\), so \(f'(x)=\frac1x\). 3. At \(x=1\), \(f(1)=3\) and \(f'(1)=1\). The tangent line is \(y-3=x-1\), or \(y=x+2\).

a) Shift the graph of \(y=\ln x\) up \(3\) units. b) \(f'(x)=\frac1x\) c) \(y=x+2\)

- Differentiate using the sum and constant multiple rules. - Recall the derivatives of sine and cosine. - Use exact unit-circle values at the two inputs. - The derivative value is the graph’s slope.

1. Differentiate: \(f'(x)=3\cos x+2\sin x\). 2. At \(x=0\), \(f'(0)=3\). 3. At \(x=\frac{\pi}{2}\), \(f'\left(\frac{\pi}{2}\right)=2\).

At \(x=0\), the slope is \(3\). At \(x=\frac{\pi}{2}\), the slope is \(2\).

- Differentiate first. - Substitute each candidate input into the derivative. - Use exact unit-circle values. - Select the inputs where the derivative equals \(1\).

1. Differentiate: \(h'(x)=\cos x-\sin x\). 2. Evaluate at the four inputs: \(h'(0)=1\), \(h'\left(\frac{\pi}{2}\right)=-1\), \(h'(\pi)=-1\), and \(h'\left(\frac{3\pi}{2}\right)=1\). Therefore, the tangent slope is \(1\) at \(x=0\) and \(x=\frac{3\pi}{2}\).

\(x=0\) and \(x=\frac{3\pi}{2}\)

- Find the derivative. - Use the range of sine. - Account for the constant factor and negative sign. - Report the values outside the derivative’s range.

1. Differentiate: \(g'(x)=-4\sin x\). 2. Since \(-1\le\sin x\le 1\), the derivative satisfies \(-4\le g'(x)\le 4\). Therefore, tangent slopes outside this interval cannot occur.

The impossible slopes are \(m<-4\) or \(m>4\).

- Recall the derivatives of sine and cosine. - A constant factor remains as a multiplier. - In part b, track both negative signs carefully.

1. For part a, use \(\frac{d}{dx}(\sin x)=\cos x\): \(f'(x)=12\cos x\). 2. For part b, use \(\frac{d}{dx}(\cos x)=-\sin x\): \(f'(x)=-\frac{3}{4}(-\sin x)=\frac{3}{4}\sin x\).

a) \(f'(x)=12\cos x\) b) \(f'(x)=\frac{3}{4}\sin x\)

- 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 \(x\ne0\)

- Recall the derivative of the sine function. - Evaluate the derivative at \(x=0\). - Use the given point and slope to write the tangent line.

1. The point lies on the graph because \(f(0)=0\). 2. Since \(f'(x)=\cos x\), the tangent slope is \(f'(0)=1\). 3. A line through the origin with slope \(1\) has equation \(y=x\).

\(y=x\)

- Differentiate both functions first. - Use slope and intercepts to identify the linear derivative. - Use the sign just to the right of \(x=0\) to distinguish the sine graphs.

1. Differentiate: \(f'(x)=2x-3\) and \(g'(x)=-2\sin x\). 2. The graph of \(f'\) is a line with slope \(2\), y-intercept \(-3\), and x-intercept \(1.5\), so it is graph (1). 3. The graph of \(g'\) is a sine curve with amplitude \(2\), reflected across the x-axis. It passes through the origin and is negative immediately to the right, so it is graph (2).

Graph (1) represents \(f'\), and graph (2) represents \(g'\).

- Identify which general-base derivative formula applies to each function before substituting the input value. - Keep the logarithm of the base in the derivative formula. - Round only after evaluating the exact derivative expression.

1. Use \(\frac{d}{dx}(a^x)=a^x\ln a\). Then \(f'(2)=3^2\ln 3=9\ln 3\approx9.888\). 2. Use \(\frac{d}{dx}(\log_a x)=\frac{1}{x\ln a}\). Then \(g'(8)=\frac{1}{8\ln 2}\approx0.180\).

a) \(9.888\) b) \(0.180\)

- Near \(x=0\), how does the height of the sine curve change as you move right? - Compare that change with the other curve's height at the same input.

1. Curve \(p\) passes through the origin and is rising near \(x=0\) with slope about \(1\), so \(p\) is the sine curve. 2. Curve \(q\) has height about \(1\) there, matching that positive slope. Thus, \(q\) is the derivative curve.

Curve \(q\). Near \(x=0\), curve \(p\) has upward slope about \(1\), while \(q(0)\approx1\).

- Use one derivative rule for each term. - Keep the negative sign in front of cosine in view. - The logarithmic term limits the domain.

1. Differentiate the terms separately: \((e^x)'=e^x\), \((-\cos x)'=\sin x\), and \((\ln x)'=1/x\). 2. Combine them to obtain \(f'(x)=e^x+\sin x+\frac{1}{x}\).

\(f'(x)=e^x+\sin x+\frac{1}{x}\), \(x>0\)

- Find the two derivatives before comparing function values. - Evaluate both rules at the same positive input. - Equal derivatives do not require equal function values.

1. For \(x>0\), \(f'(x)=\frac{1}{x}\) and \(g'(x)=\frac{1}{2\sqrt{x}}\). 2. At \(x=4\), these slopes are \(\frac{1}{4}\) and \(\frac{1}{4}\). 3. The tangent slopes are equal, although the function values differ.

Yes. Both tangent slopes at \(x=4\) equal \(\frac{1}{4}\).

- The derivative gives the tangent slope. - Parallel lines have equal slopes; perpendicular nonvertical lines have slopes whose product is \(-1\). - Use unit-circle values and periodicity. - Substitute each x-coordinate into the original function.

1. The derivative is \(f'(x)=-\sin x\). 2. In part a, a parallel tangent must have slope \(-\frac{1}{2}\). Thus, \(-\sin x=-\frac{1}{2}\), so \(\sin x=\frac{1}{2}\). The solutions are \(x=\frac{\pi}{6}+2k\pi\) or \(x=\frac{5\pi}{6}+2k\pi\), where \(k\in\mathbb{Z}\). The corresponding y-coordinates are \(\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\). 3. In part b, a line perpendicular to \(y=x\) has slope \(-1\). Thus, \(-\sin x=-1\), so \(\sin x=1\). Therefore, \(x=\frac{\pi}{2}+2k\pi\), and the y-coordinate is \(0\).

a) \(\left(\frac{\pi}{6}+2k\pi, \frac{\sqrt{3}}{2}\right)\) and \(\left(\frac{5\pi}{6}+2k\pi, -\frac{\sqrt{3}}{2}\right)\), where \(k\in\mathbb{Z}\) b) \(\left(\frac{\pi}{2}+2k\pi, 0\right)\), where \(k\in\mathbb{Z}\)

- Differentiate using the constant multiple rule. - Set the derivative equal to the given slope. - Use the unit circle and periodicity to solve the trigonometric equation. - Substitute the inputs into the original function for the y-coordinates.

1. Differentiate: \(g'(x)=2\cos x\). 2. Set the derivative equal to the required slope: \(2\cos x=\sqrt{2}\), so \(\cos x=\frac{\sqrt{2}}{2}\). Therefore, \(x=\frac{\pi}{4}+2k\pi\) or \(x=-\frac{\pi}{4}+2k\pi\), where \(k\in\mathbb{Z}\). 3. Evaluate \(g\): \(g\left(\frac{\pi}{4}\right)=\sqrt{2}\) and \(g\left(-\frac{\pi}{4}\right)=-\sqrt{2}\).

\(\left(\frac{\pi}{4}+2k\pi, \sqrt{2}\right)\) and \(\left(-\frac{\pi}{4}+2k\pi, -\sqrt{2}\right)\), where \(k\in\mathbb{Z}\)

- The derivative gives the tangent slope. - Use the exact cosine value at \(\frac{2\pi}{3}\). - Write the line in point-slope form. - Keep all angle measures in radians.

1. Differentiate: \(f'(x)=\cos x\). 2. The tangent slope is \(m=f'\left(\frac{2\pi}{3}\right)=\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}\). 3. Use point-slope form: \(y-\frac{\sqrt{3}}{2}=-\frac{1}{2}\left(x-\frac{2\pi}{3}\right)\). Simplifying gives \(y=-\frac{1}{2}x+\frac{\pi}{3}+\frac{\sqrt{3}}{2}\).

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

- Evaluate the function to find the point of tangency. - Differentiate using the constant multiple rule. - Use the exact sine and cosine values at \(\frac{\pi}{6}\). - Write the tangent line in point-slope form.

1. Find the point of tangency: \(g\left(\frac{\pi}{6}\right)=2\cos\left(\frac{\pi}{6}\right)=\sqrt{3}\). 2. Differentiate: \(g'(x)=-2\sin x\). The tangent slope is \(g'\left(\frac{\pi}{6}\right)=-2\cdot\frac{1}{2}=-1\). 3. Use point-slope form: \(y-\sqrt{3}=-\left(x-\frac{\pi}{6}\right)\). Therefore, \(y=-x+\frac{\pi}{6}+\sqrt{3}\).

\(y=-x+\frac{\pi}{6}+\sqrt{3}\)

- Evaluate the function and derivative at the given input. - Use point-slope form for the tangent line. - The derivative value is the tangent slope. - Choose the angle of inclination in the interval \([0^\circ, 180^\circ)\).

1. Evaluate the function: \(f\left(\frac{\pi}{2}\right)=0-2=-2\). Thus, the point is \(\left(\frac{\pi}{2}, -2\right)\). 2. Differentiate: \(f'(x)=-\sin x-2\cos x\). The tangent slope is \(f'\left(\frac{\pi}{2}\right)=-1\). 3. Use point-slope form: \(y+2=-\left(x-\frac{\pi}{2}\right)\). Therefore, \(y=-x+\frac{\pi}{2}-2\). 4. The angle of inclination is measured counterclockwise from the positive x-axis and satisfies \(\tan\alpha=-1\). Thus, \(\alpha=135^\circ\).

Tangent line: \(y=-x+\frac{\pi}{2}-2\) Angle of inclination: \(135^\circ\)

- Set the two function values equal to find the intersection. - Use the sine and cosine derivative formulas for the tangent slopes. - Convert each slope to its inclination angle in \([0^\circ,180^\circ)\). - Use the smaller difference of the inclination angles and the requested precision.

1. At an intersection, \(2\sin x=2\cos x\). On \([0,\pi]\), this gives \(x=\frac{\pi}{4}\), with y-coordinate \(\sqrt{2}\). Thus, \(S=\left(\frac{\pi}{4},\sqrt{2}\right)\). 2. The derivatives are \(f'(x)=2\cos x\) and \(g'(x)=-2\sin x\). At \(S\), the slopes are \(\sqrt{2}\) and \(-\sqrt{2}\). 3. Therefore, \(\alpha=\arctan(\sqrt{2})\approx54.74^\circ\) and \(\beta=180^\circ-54.74^\circ\approx125.26^\circ\). 4. The smaller intersection angle is \(\gamma=\beta-\alpha\approx70.53^\circ\).

a) \(S=\left(\frac{\pi}{4},\sqrt{2}\right)\) b) \(\alpha\approx54.74^\circ\); \(\beta\approx125.26^\circ\) c) \(\gamma\approx70.53^\circ\)

- Evaluate the function and derivative at the given input. - Use point-slope form. - Relate the derivative value to the tangent slope. - Give the inclination angle in \([0^\circ, 180^\circ)\).

1. The point of tangency is \(\left(\frac{\pi}{2}, 1\right)\). 2. Differentiate: \(f'(x)=\cos x-\sin x\). The slope at \(x=\frac{\pi}{2}\) is \(-1\). 3. Use point-slope form: \(y-1=-\left(x-\frac{\pi}{2}\right)\). Therefore, \(y=-x+\frac{\pi}{2}+1\). 4. Since the slope is \(-1\), the angle of inclination is \(135^\circ\).

Tangent line: \(y=-x+\frac{\pi}{2}+1\) Angle of inclination: \(135^\circ\)

- Evaluate the function to find the point. - Differentiate cosine and keep the constant factor. - Use point-slope form for the tangent line. - A slope of \(-\sqrt{3}\) has an inclination angle in the second quadrant.

1. Evaluate the function: \(g\left(\frac{2\pi}{3}\right)=2\cdot\left(-\frac{1}{2}\right)+1=0\). Thus, the point is \(\left(\frac{2\pi}{3}, 0\right)\). 2. Differentiate: \(g'(x)=-2\sin x\). The tangent slope is \(g'\left(\frac{2\pi}{3}\right)=-\sqrt{3}\). 3. Use point-slope form: \(y=-\sqrt{3}\left(x-\frac{2\pi}{3}\right)\). Therefore, \(y=-\sqrt{3}x+\frac{2\pi\sqrt{3}}{3}\). 4. The standard angle of inclination in \([0^\circ, 180^\circ)\) is \(120^\circ\).

Tangent line: \(y=-\sqrt{3}x+\frac{2\pi\sqrt{3}}{3}\) Angle of inclination: \(120^\circ\)

- Evaluate the function at the given input. - Use the derivative to find the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Write both lines in point-slope form.

1. Find the point: \(f(\pi)=2\cos\pi+\pi=\pi-2\). Thus, \(P=(\pi, \pi-2)\). 2. Differentiate: \(f'(x)=-2\sin x+1\). The tangent slope is \(f'(\pi)=1\). Therefore, the tangent line is \(y-(\pi-2)=x-\pi\), or \(y=x-2\). 3. The normal slope is \(-1\). Thus, \(y-(\pi-2)=-(x-\pi)\), or \(y=-x+2\pi-2\).

Tangent line: \(y=x-2\) Normal line: \(y=-x+2\pi-2\)

- Evaluate the function to locate the point. - Differentiate sine and cosine carefully. - Use the derivative value as the tangent slope. - Use the negative reciprocal for the normal slope.

1. Evaluate the function: \(f\left(\frac{\pi}{2}\right)=1\). Thus, the point is \(\left(\frac{\pi}{2}, 1\right)\). 2. Differentiate: \(f'(x)=\cos x+\sin x\). The tangent slope is \(f'\left(\frac{\pi}{2}\right)=1\). Therefore, the tangent line is \(y-1=x-\frac{\pi}{2}\), or \(y=x-\frac{\pi}{2}+1\). 3. The normal slope is \(-1\). Thus, \(y-1=-\left(x-\frac{\pi}{2}\right)\), or \(y=-x+\frac{\pi}{2}+1\).

Tangent line: \(y=x-\frac{\pi}{2}+1\) Normal line: \(y=-x+\frac{\pi}{2}+1\)

- What is the point-slope formula for a tangent line at \(x=x_0\)? - What form does an equation of a line through the origin have? - Recall the derivative rule for \(a^x\). - Compare the slope of \(e^x\) at each point with the function value there.

1. Since \(f(0)=e^0=1\) and \(f'(x)=e^x\), the slope at \(x=0\) is \(f'(0)=1\). Using point-slope form, \(y-1=1(x-0)\), so the tangent line is \(y=x+1\). 2. At \(x=1\), \(f(1)=e\) and \(f'(1)=e\). The tangent line is \(y-e=e(x-1)\), which simplifies to \(y=ex\). Because this line contains \((0, 0)\), it passes through the origin. 3. For \(f(x)=e^x\), the derivative is the function itself: \(f'(x)=e^x=f(x)\). For a different base, \(a^x\), the derivative is \(\ln(a)a^x\). Only when \(a=e\) is the factor \(\ln(a)\) equal to \(1\).

a) \(y=x+1\) b) The tangent line is \(y=ex\), so it passes through \((0, 0)\). c) The base \(e\) is the unique positive base for which the derivative of \(e^x\) equals \(e^x\) itself.

- Substitute each value of \(a\) and the given value of \(h\) carefully into the difference quotient. - Compare both approximate slopes with \(1\). - Recall the derivative rule for \(a^x\) and consider what happens when \(\ln(a)=1\). - Which differentiation rule applies to \(e^{2x+5}\)?

1. For \(a=2\), \(\frac{2^{0.0001}-1}{0.0001}\approx0.6932\). 2. For \(a=3\), \(\frac{3^{0.0001}-1}{0.0001}\approx1.0987\). 3. In general, \(\frac{d}{dx}a^x=\ln(a)a^x\). Because \(\ln(e)=1\), differentiating an exponential with base \(e\) does not introduce an additional base-dependent factor. Applying the chain rule gives \(f'(x)=2e^{2x+5}\).

a) For base \(2\), the slope is approximately \(0.6932\). For base \(3\), the slope is approximately \(1.0987\). b) Since \(\frac{d}{dx}e^x=e^x\), only the derivative of the exponent contributes an additional factor. Thus, \(f'(x)=2e^{2x+5}\).

- Find both \(f(-1)\) and \(f'(-1)\). - Use point-slope form for the tangent line. - Keep \(e\) symbolic in the exact equation. - Use the inverse tangent of the slope to find the angle.

1. Find the point of tangency: \(f(-1)=10e^{-1}+1=\frac{10}{e}+1\). 2. Differentiate: \(f'(x)=10e^x+2x\). The slope at \(x=-1\) is \(m=f'(-1)=\frac{10}{e}-2\). 3. Use point-slope form: \(y-\left(\frac{10}{e}+1\right)=\left(\frac{10}{e}-2\right)(x+1)\). Simplifying gives \(y=\left(\frac{10}{e}-2\right)x+\frac{20}{e}-1\). 4. The angle of inclination satisfies \(\tan\theta=m\). Therefore, \(\theta=\arctan\left(\frac{10}{e}-2\right)\approx 59.2^\circ\).

Tangent line: \(y=\left(\frac{10}{e}-2\right)x+\frac{20}{e}-1\) Angle of inclination: \(\theta\approx 59.2^\circ\)

- Use the basic derivative formulas for \(e^x\) and \(\ln x\). - Check a claimed intercept by substitution. - Connect a positive first derivative with increasing behavior. - Treat a vertical translation as a change in output values, not in derivative structure.

1. The constant-multiple rule gives \(g'(x)=3e^x\), so statement a is true. 2. Since \(h(0)=e^0-1=0\), the graph contains \((0,0)\). Statement b is true. 3. Since \(f'(x)=e^x>0\) for every real \(x\), \(f\) is strictly increasing. Statement c is true. 4. For \(x>0\), \(q'(x)=\frac{1}{x}\), not \(\ln x\). Statement d is false. 5. The range of \(e^x\) is \((0,\infty)\). Shifting down \(5\) units gives \((-5,\infty)\). Statement e is true.

a) True b) True c) True d) False e) True

- Use the point-slope form of a tangent line. - Find both the function value and derivative value at \(x=0\). - A point on the tangent line must satisfy its equation. - Express the tangent slope and y-intercept in terms of \(k\).

1. Differentiate: \(f_k'(x)=ke^x-1\). 2. At \(x=0\), \(f_k(0)=k\) and \(f_k'(0)=k-1\). 3. The tangent line is \(y=(k-1)x+k\). 4. Substitute \((2, 5)\): \(5=2(k-1)+k\). 5. Solve \(5=3k-2\) to obtain \(k=\frac{7}{3}\).

a) \(f_k'(x)=ke^x-1\) b) \(k=\frac{7}{3}\)

- For consecutive integer inputs, compare the two outputs by division. - Use either point to find the scale factor after finding the base. - Rewriting the exponential with base \(e\) makes the exponent's derivative visible.

1. The output rises from \(45\) to \(135\), so the function shows exponential growth. The ratio for one input step is \(a=135/45=3\). 2. From \(45=ba\), obtain \(b=15\). Thus \(f(x)=15\cdot3^x\). 3. Since \(3=e^{\ln(3)}\), we get \(f(x)=15e^{x\ln(3)}\), so \(k=\ln(3)\). The chain rule gives \(f'(x)=15\ln(3)e^{x\ln(3)}\). 4. At zero, the slope is \(f'(0)=15\ln(3)\approx16.48\).

a) Growth, because the output increases as the input increases. b) \(a=3\), \(b=15\). c) \(f(x)=15e^{x\ln(3)}\); \(f'(x)=15\ln(3)e^{x\ln(3)}\); \(f'(0)=15\ln(3)\approx16.48\).

- Write the tangent line at a general input \(u\). - A line through the origin has y-intercept \(0\). - The tangent slope is the derivative value at the point of tangency. - Use logarithm properties for powers of \(e\).

1. Differentiate: \(f'(x)=\frac{1}{x}\). 2. At \(x_0=e^2\), \(f(e^2)=2\) and \(f'(e^2)=\frac{1}{e^2}\). Therefore, \(y=\frac{1}{e^2}(x-e^2)+2=\frac{x}{e^2}+1\). 3. The tangent line at a general input \(u>0\) is \(y=\frac{1}{u}(x-u)+\ln u=\frac{x}{u}+\ln u-1\). For this line to pass through the origin, \(\ln u-1=0\), so \(u=e\). The point of tangency is \((e, 1)\), and the tangent line is \(y=\frac{x}{e}\). 4. At \(x_1=\sqrt{e}\), the tangent line is \(y=\frac{x}{\sqrt{e}}-\frac{1}{2}\). Setting \(y=0\) gives the x-intercept \(x=\frac{\sqrt{e}}{2}\).

a) \(y=\frac{x}{e^2}+1\) b) Point of tangency: \((e, 1)\); tangent line: \(y=\frac{x}{e}\) c) x-intercept: \(x=\frac{\sqrt{e}}{2}\)

- Parallel lines have equal slopes. - Write the tangent line at a general input \(u\). - A point on the y-axis determines the tangent’s y-intercept. - Set \(x=0\) in the general tangent equation.

1. A line parallel to \(y=x\) has slope \(1\). Since \(f'(x)=\frac{1}{x}\), solve \(\frac{1}{x}=1\), giving \(x=1\). Since \(f(1)=0\), the tangent line is \(y=x-1\). 2. The tangent line at a general input \(u>0\) is \(y=\frac{x}{u}+\ln u-1\). Passing through \((0, -2)\) requires \(\ln u-1=-2\), so \(u=e^{-1}\). The slope is \(1/u=e\), giving \(y=ex-2\). 3. From the general tangent equation, setting \(x=0\) gives \(y=\ln u-1\), so the y-intercept is \((0, \ln u-1)\). For an intercept of \(3\), \(\ln u-1=3\), so \(u=e^4\).

a) \(y=x-1\) b) \(y=ex-2\) c) The y-intercept is \((0, \ln u-1)\), and \(u=e^4\) when the intercept is \(3\).

- The derivative gives the slope of the graph. - Check any solution against the function's domain. - Write the tangent line at a general x-value \(u\), then use the fact that it passes through the origin. - Recall that \(\ln(x)\) is defined only for positive \(x\).

1. Differentiate: \(f'(x)=3+\frac{2}{x}\). 2. For slope \(5\), solve \(3+\frac{2}{x}=5\). This gives \(x=1\), and \(f(1)=3+2\ln(1)=3\). The point is \(P(1, 3)\). 3. For slope \(2\), solve \(3+\frac{2}{x}=2\), which gives \(x=-2\). This value is outside the domain \(x>0\), so no such point exists. 4. Let the tangent point have x-coordinate \(u>0\). The tangent line is \(y=f'(u)(x-u)+f(u)\). Because it passes through \((0, 0)\), \(0=\left(3+\frac{2}{u}\right)(-u)+3u+2\ln(u)\). 5. Simplifying gives \(-2+2\ln(u)=0\), so \(\ln(u)=1\) and \(u=e\). 6. The slope is \(f'(e)=3+\frac{2}{e}\). Since the line passes through the origin, its equation is \(y=\left(3+\frac{2}{e}\right)x\).

a) \(P(1, 3)\) b) No. Solving for slope \(2\) gives \(x=-2\), which is outside the domain. c) \(y=\left(3+\frac{2}{e}\right)x\)

- Write the tangent line at a general x-value \(x_0\). - Substitute the coordinates of \(P\) into that line. - Factor the resulting equation. - Explain why the exponential factor cannot be zero.

1. Differentiate: \(f_k'(x)=\frac{1}{k}e^x\). 2. The tangent line at \(x=x_0\) is \(y=f_k'(x_0)(x-x_0)+f_k(x_0)\). 3. Requiring the line to pass through \(P(k, 0)\) gives \(0=\frac{1}{k}e^{x_0}(k-x_0)+\frac{1}{k}e^{x_0}\). 4. Factor: \(0=\frac{1}{k}e^{x_0}(k-x_0+1)\). Since \(k>0\) and \(e^{x_0}>0\), the nonzero factor can be divided out. Therefore, \(x_0=k+1\). This solution is unique because the remaining equation is linear. 5. Substitute \(x_0=k+1\): \(y=\frac{e^{k+1}}{k}(x-k-1)+\frac{e^{k+1}}{k}=\frac{e^{k+1}}{k}(x-k)\).

The point of tangency occurs at \(x_0=k+1\), and the tangent line is \(y=\frac{e^{k+1}}{k}(x-k)\).

- Recall the derivative of \(\ln(x)\). - Express a percent increase as a multiplication factor. - Expand \(\ln(kx)\) using the product property. - Check whether the original input cancels from the difference.

1. The derivative is \(f'(x)=\frac{a}{x}\). The condition \(f'(2)=1.5\) gives \(\frac{a}{2}=1.5\), so \(a=3\). 2. A \(20\%\) increase changes \(x\) to \(1.2x\). Thus, \(f(1.2x)-f(x)=3\ln(1.2x)-3\ln(x)=3\ln(1.2)\approx0.547\). 3. A \(p\%\) increase changes \(x\) to \(x\left(1+\frac{p}{100}\right)\). The change is \(a\ln\!\left(x\left(1+\frac{p}{100}\right)\right)-a\ln(x)=a\ln\!\left(1+\frac{p}{100}\right)\), which is independent of \(x\).

1. \(a=3\) 2. \(3\ln(1.2)\approx0.547\) 3. \(a\ln\!\left(1+\frac{p}{100}\right)\)

- Read the effects of the factor and constant from the formula. - Parallel lines have equal slopes. - Set the derivative equal to the target slope. - Substitute the resulting input into the original function.

1. Multiply the y-values of \(\ln x\) by \(2\), reflect the graph across the x-axis, and then shift it up \(1\) unit. 2. The given line has slope \(-1\). Differentiate: \(g'(x)=-\frac{2}{x}\). Set the derivative equal to \(-1\): \(-\frac{2}{x}=-1\), so \(x=2\). The corresponding y-value is \(g(2)=1-2\ln2\).

a) Vertically stretch by a factor of \(2\), reflect across the x-axis, and shift up \(1\) unit. b) \((2, 1-2\ln2)\)

- Substitute \(x=1\). - Differentiate with respect to \(x\). - Set two family members equal and cancel their common terms. - Solve the remaining logarithmic equation.

1. \(f_k(1)=k\ln(1)-1+1=0\), so every graph passes through \((1, 0)\). 2. The derivative is \(f_k'(x)=\frac{k}{x}-1\). Thus, the slope at \(x=1\) is \(f_k'(1)=k-1\). 3. If two graphs intersect, then \(k_1\ln(x)-x+1=k_2\ln(x)-x+1\). Hence, \((k_1-k_2)\ln(x)=0\). Since \(k_1\neq k_2\), \(\ln(x)=0\), so \(x=1\). The corresponding output is \(0\), so the only shared point is \((1, 0)\).

1. \(f_k(1)=0\) 2. \(k-1\) 3. The only shared point is \((1, 0)\).

- Recall the derivative of \(\log_bx\). - Find the function value and derivative value at \(x=3\). - Use point-slope form. - Convert between base-3 and natural logarithms if needed.

1. The point on the graph is \(f(3)=\log_33=1\). 2. Differentiate: \(f'(x)=\frac{1}{x\ln3}\). Thus, the slope at \(x=3\) is \(f'(3)=\frac{1}{3\ln3}\). 3. Use point-slope form: \(y-1=\frac{1}{3\ln3}(x-3)\). Therefore, \(y=\frac{x}{3\ln3}+1-\frac{1}{\ln3}\).

\(y=\frac{x}{3\ln3}+1-\frac{1}{\ln3}\)

- Recall the derivative of \(\log_bx\). - Substitute the given input and slope. - Solve the resulting equation for \(\ln b\). - Exponentiate to solve for \(b\).

1. Differentiate: \(h'(x)=\frac{1}{x\ln b}\). 2. Use the given slope: \(\frac{1}{2\ln b}=0.5=\frac{1}{2}\). Therefore, \(\ln b=1\), so \(b=e\).

\(b=e\)

- Tangent graphs have the same function value and slope at the point of tangency. - Use the logarithm derivative rule to determine the required slope at \(x=1\). - Start with a general quadratic and translate the two tangency conditions plus the extra point into three equations.

1. Write \(p(x)=ax^2+bx+c\). 2. Tangency at \((1,0)\) requires \(p(1)=0\) and \(p'(1)=f'(1)\). 3. Since \(f'(x)=\frac1x\), \(f'(1)=1\). Therefore, \(a+b+c=0\) and \(2a+b=1\). 4. The point \(A(2,4)\) gives \(4a+2b+c=4\). 5. Solving the three equations gives \(a=3\), \(b=-5\), and \(c=2\). Thus \(p(x)=3x^2-5x+2\).

\(p(x)=3x^2-5x+2\)

- Rewrite the function to make the transformations visible. - Consider what repeated multiplication by \(0.85\) does as \(t\) increases. - Use the derivative rule for an exponential function with base \(a\). - What does a rate at \(t=0\) describe about the start of the process?

1. Rewrite the function as \(N(t)=-800(0.85^t)+800\). Starting with \(h(t)=0.85^t\), reflect the graph across the \(t\)-axis, stretch it vertically by a factor of \(800\), and shift it up \(800\) units. 2. Since \(0<0.85<1\), \(0.85^t\to 0\) as \(t\to\infty\). Therefore, \(N(t)\to 800\), meaning the model predicts that eventually all \(800\) employees will know the announcement. 3. Using \(\frac{d}{dt}(a^t)=a^t\ln(a)\), \(N'(t)=-800\ln(0.85)(0.85^t)\). 4. At \(t=0\), \(N'(0)=-800\ln(0.85)\approx 130.02\). Initially, the announcement is spreading at about \(130\) employees per hour.

a) Reflect across the \(t\)-axis, stretch vertically by a factor of \(800\), and shift up \(800\) units. b) \(N(t)\to 800\). The model predicts that the announcement eventually reaches the entire company. c) \(N'(t)=-800\ln(0.85)(0.85^t)\), and \(N'(0)\approx 130.02\) employees per hour.

- Determine the values of cosine on the stated interval. - Use point-slope form for the tangent line. - Find both axis intercepts of the tangent line. - Use the area formula for a right triangle.

1. On the given interval, \(0\leq\cos(x)\leq1\). Since \(k>0\), the range is \([0, k]\). 2. Since \(h_k\left(\frac{\pi}{4}\right)=\frac{k\sqrt{2}}{2}\) and \(h_k'(x)=-k\sin(x)\), the tangent slope is \(-\frac{k\sqrt{2}}{2}\). Thus \(t_k: y=-\frac{k\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)+\frac{k\sqrt{2}}{2}\). 3. The y-intercept is \(\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)\), and the x-intercept is \(1+\frac{\pi}{4}\). Therefore, \(A(k)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)=\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2\). 4. Set \(A(k)=10\): \(\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2=10\). Solving gives \(k=\frac{20\sqrt{2}}{\left(1+\frac{\pi}{4}\right)^2}\approx8.87\).

1. \([0, k]\) 2. \(t_k: y=-\frac{k\sqrt{2}}{2}x+\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)\) 3. \(A(k)=\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2\) 4. \(k=\frac{20\sqrt{2}}{\left(1+\frac{\pi}{4}\right)^2}\approx8.87\)

- Compare the two graph values at \(x=0\). - Match the sign of the derivative with where cosine is increasing or decreasing. - Use a phase-shift identity relating sine and cosine.

1. Since \(f(0)=1\), the solid curve, which passes through \((0,1)\), represents \(f\). At \(x=0\), cosine has a local maximum, so its derivative is \(0\). The dashed curve passes through \((0,0)\) and is negative on \((0,\pi)\), matching \(f'(x)=-\sin x\). 2. Since \(\cos\left(x+\frac{\pi}{2}\right)=-\sin x\), the smallest positive value is \(c=\frac{\pi}{2}\). This shifts the cosine graph left by \(\frac{\pi}{2}\).

a) The solid curve represents \(f(x)=\cos x\), and the dashed curve represents \(f'(x)=-\sin x\). b) \(c=\frac{\pi}{2}\). Shift the graph of \(f\) left by \(\frac{\pi}{2}\).

- Compare the three formulas term by term. - A tangent line requires both \(f(x_0)\) and \(f'(x_0)\). - A horizontal tangent has derivative value \(0\).

1. Since \(f(x)=g(x)+h(x)\), the y-coordinate on \(f\) is the sum of the corresponding y-coordinates on \(g\) and \(h\). At \(x=0\), \(f(0)=1\). 2. Differentiate: \(f'(x)=0.5-\sin x\). At \(x=\pi\), \(f'(\pi)=0.5\) and \(f(\pi)=\frac{\pi}{2}-1\). Thus, the tangent is \(t(x)=0.5x-1\). 3. Horizontal tangents satisfy \(\sin x=\frac{1}{2}\). On \([0,2\pi]\), the solutions are \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\).

a) Add corresponding y-values: \(f(x)=g(x)+h(x)\). At \(x=0\), \(f(0)=1\). b) \(t(x)=0.5x-1\) c) \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)

- Use point-slope form for a tangent line at \(x=a\). - Recall the derivative of the natural logarithm. - A y-intercept occurs where \(x=0\). - Subtract the two y-coordinates and simplify. - A result with no \(a\) is constant for all allowed values of \(a\).

1. Differentiate: \(f'(x)=\frac{1}{x}\), so the tangent slope at \(x=a\) is \(\frac{1}{a}\). 2. Use point-slope form: \(y-\ln(a)=\frac{1}{a}(x-a)\). Therefore, \(t: y=\frac{1}{a}x-1+\ln(a)\). 3. Set \(x=0\) to find the y-intercept: \(y_S=\ln(a)-1\). Thus \(S=(0, \ln(a)-1)\). 4. The y-coordinate of \(P\) is \(\ln(a)\). The vertical distance is \(\ln(a)-(\ln(a)-1)=1\), which is independent of \(a\).

a) \(t: y=\frac{1}{a}x-1+\ln(a)\) b) \(S=(0, \ln(a)-1)\) c) The constant vertical distance is \(1\).

- Determine where sine and cosine have the same value on the stated interval. - Differentiate both functions to find the tangent slopes at the intersection. - Use the acute angle-between-lines formula and the requested precision.

1. Solve \(\sin x=\cos x\) on \([0,\pi]\), giving \(x=\frac{\pi}{4}\). The intersection is \(\left(\frac{\pi}{4},\frac{\sqrt{2}}{2}\right)\). 2. The derivatives are \(f'(x)=\cos x\) and \(g'(x)=-\sin x\). At the intersection, the slopes are \(\frac{\sqrt{2}}{2}\) and \(-\frac{\sqrt{2}}{2}\). 3. Thus, \(\tan(\alpha)=2\sqrt{2}\), so \(\alpha=\arctan(2\sqrt{2})\approx70.5^\circ\).

The intersection is \(\left(\frac{\pi}{4},\frac{\sqrt{2}}{2}\right)\), and the acute intersection angle is approximately \(70.5^\circ\).

- Identify where the sine curve has horizontal tangents. - Determine where it is increasing or decreasing most steeply. - Match the resulting periodic slope pattern to a familiar trig graph.

1. The slopes are \(f'(0)=1\), \(f'\left(\frac{\pi}{2}\right)=0\), \(f'(\pi)=-1\), \(f'\left(\frac{3\pi}{2}\right)=0\), and \(f'(2\pi)=1\). 2. These slope values produce the cosine graph. 3. Therefore, \(f'(x)=\cos x\).

\(f'(0)=1\), \(f'\left(\frac{\pi}{2}\right)=0\), \(f'(\pi)=-1\), \(f'\left(\frac{3\pi}{2}\right)=0\), \(f'(2\pi)=1\); the derivative is \(\cos x\).

- Add corresponding y-values of the two component functions. - Parallel lines have equal slopes. - Use the bound \(-1\le\sin x\le1\) for the final positivity argument.

1. Since \(f(x)=g(x)+h(x)\), the graph of \(f\) is obtained by adding the corresponding y-values, so it oscillates around \(y=x\). 2. The line \(g\) has slope \(1\). Since \(f'(x)=1+1.5\cos x\), parallel tangents satisfy \(\cos x=0\), giving \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) on the stated interval. 3. For \(0<x\le\pi\), \(\sin x\ge0\), so \(f(x)>0\). For \(x>\pi\), \(f(x)\ge x-1.5>\pi-1.5>0\).

a) The graph of \(f\) is the pointwise sum of \(g\) and \(h\), so it oscillates around \(y=x\). b) \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) c) For \(0<x\le\pi\), both terms are nonnegative. For \(x>\pi\), \(f(x)\ge x-1.5>0\).

- Compare the shapes and formulas of the three functions. - Set the derivative equal to the desired slope in part b. - A horizontal tangent requires the derivative to equal \(0\). - Use the unit circle for the resulting sine equations.

1. Graph B is the line through the origin, so it represents \(g\). Graph C is the reflected cosine curve, so it represents \(h\). Their pointwise sum is graph A, which represents \(f\). 2. Since \(f'(x)=0.5+\sin x\), slope \(0.5\) occurs when \(\sin x=0\). On \([0,2\pi]\), the inputs are \(0\), \(\pi\), and \(2\pi\), giving points \((0,-1)\), \(\left(\pi,\frac{\pi}{2}+1\right)\), and \((2\pi,\pi-1)\). 3. Horizontal tangents satisfy \(\sin x=-\frac{1}{2}\), so \(x=\frac{7\pi}{6}+2\pi n\) or \(x=\frac{11\pi}{6}+2\pi n\), where \(n\in\mathbb{Z}\).

a) A: \(f\); B: \(g\); C: \(h\) b) \((0,-1)\), \(\left(\pi,\frac{\pi}{2}+1\right)\), and \((2\pi,\pi-1)\) c) \(x=\frac{7\pi}{6}+2\pi n\) or \(x=\frac{11\pi}{6}+2\pi n\), where \(n\in\mathbb{Z}\)

- Differentiate the natural logarithm. - Substitute \(x=0\) into the general tangent equation. - Subtract the y-coordinate of \(S\) from \(f(a)\). - Use the displayed intercepts to find the tangent slope, then set it equal to \(f^{\prime}(a)\).

1. Since \(f^{\prime}(x)=\frac{1}{x}\), the tangent at \(x=a\) has equation \(y-\ln(a)=\frac{1}{a}(x-a)\). 2. Simplifying gives \(y=\frac{1}{a}x-1+\ln(a)\). At \(x=0\), \(y=\ln(a)-1\), so the y-intercept is \(S(0, \ln(a)-1)\). 3. The vertical difference is \(f(a)-y_S=\ln(a)-(\ln(a)-1)=1\). 4. In the displayed example, the tangent passes through \((0, -1)\) and \((1, 0)\), so its slope is \(1\). Because the tangent slope is \(\frac{1}{a}\), \(\frac{1}{a}=1\), and therefore \(a=1\).

a) The tangent is \(y=\frac{1}{a}x-1+\ln(a)\), so its y-intercept is \(S(0, \ln(a)-1)\). b) The vertical distance is always \(1\) unit. c) \(a=1\)

- Evaluate the graphs at \(x=1\). - Solve the zero equation for \(x\). - The derivative of an added constant is zero.

1. Since \(\ln(1)=0\), \(f_c(1)=c\). Graph p has value \(2\) at \(x=1\), so \(c=2\). Graph q has value \(-1\) at \(x=1\), so \(c=-1\). 2. A zero satisfies \(\ln(x)+c=0\), so \(x=e^{-c}\). As \(c\) increases, \(-c\) decreases, so the zero moves left toward the y-axis. 3. The derivative is \(f_c'(x)=\frac{1}{x}\), which does not depend on \(c\). Therefore, every graph has slope \(f_c'(2)=\frac{1}{2}\) at \(x=2\).

a) p: \(c=2\); q: \(c=-1\) b) The zero is \(x=e^{-c}\) and moves left as \(c\) increases. c) \(f_c'(x)=\frac{1}{x}\), so the common slope at \(x=2\) is \(\frac{1}{2}\).

- Parallel lines have equal slopes. - Use the double-angle identity for sine. - Perpendicular nonvertical lines have a slope product of \(-1\). - Use the maximum possible values of sine and cosine in part d.

1. The derivatives are \(f'(x)=\cos x\) and \(g'(x)=-\sin x\). Parallel tangents have equal slopes, so \(\cos x=-\sin x\). On \([0, \pi]\), this gives \(x_P=\frac{3\pi}{4}\). 2. The product of the slopes is \(f'(x)g'(x)=-\sin x\cos x=-\frac{1}{2}\sin(2x)\). 3. Perpendicular nonvertical lines require a slope product of \(-1\). That would require \(-\frac{1}{2}\sin(2x)=-1\), or \(\sin(2x)=2\), which is impossible. Therefore, no common input produces perpendicular tangents. 4. For different inputs, require \(\cos(x_1)[-\sin(x_2)]=-1\), so \(\cos(x_1)\sin(x_2)=1\). On \([0, \pi]\), this occurs when \(x_1=0\) and \(x_2=\frac{\pi}{2}\).

a) \(x_P=\frac{3\pi}{4}\) b) \(f'(x)g'(x)=-\frac{1}{2}\sin(2x)\) c) No such common input exists. d) \(x_1=0\), \(x_2=\frac{\pi}{2}\)