Free Math Worksheets

- 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 the polynomial and trigonometric terms separately. - Track the signs when differentiating sine and cosine. - Compare each result with the choices. - Each derivative choice is used once.

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

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

- Think about the end behavior of the basic exponential graph. - What values can \(e^x\) take? - Recall the derivative of the natural exponential function. - What algebraic condition characterizes origin symmetry? - Compare a horizontal change inside the exponent with a vertical change outside the exponential expression.

1. As \(x\to-\infty\), \(e^x\to0\), so statement a is true. 2. Since \(e^x>0\) for every real \(x\), \(e^x+2>2\). Therefore, \(g\) has no real zeros, so statement b is false. 3. Because \(f'(x)=e^x\), the slope at \(x=0\) is \(f'(0)=e^0=1\). Statement c is true. 4. Origin symmetry would require \(f(-x)=-f(x)\), but \(e^{-x}\ne-e^x\). Statement d is false. 5. Replacing \(x\) with \(x-2\) shifts the graph \(2\) units right. Statement e is true.

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

- The argument of a logarithm must be positive. - Use the product property of logarithms. - Move an exponent in front of a logarithm. - Recall the derivative of \(\ln x\).

1. The logarithm requires \(x^5e^3>0\). Since \(e^3>0\), this is equivalent to \(x^5>0\), so \(x>0\). Therefore, \(D_f=(0, \infty)\). 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) \(D_f=(0, \infty)\) b) \(f(x)=5\ln x+3\) c) \(f'(x)=\frac{5}{x}\)

- Determine when \(5x^4\) is positive. - Split the logarithm of a product into a sum. - Use absolute value when rewriting \(\ln(x^4)\) for a domain containing negative inputs. - Recall the derivative of \(\ln|x|\).

1. The logarithm requires \(5x^4>0\). Since \(x^4>0\) exactly when \(x\ne0\), \(D_f=\mathbb{R}\setminus\{0\}\). 2. Use the product and power properties, accounting for negative values of \(x\): \(f(x)=\ln5+\ln(x^4)=\ln5+4\ln|x|\). 3. Differentiate on the domain: \(f'(x)=\frac{4}{x}\).

1. \(D_f=\mathbb{R}\setminus\{0\}\) 2. \(f(x)=\ln5+4\ln|x|\) 3. \(f'(x)=\frac{4}{x}\)

- Combine the square-root and logarithm domain restrictions. - Rewrite the square root as a fractional power. - Use the quotient property of logarithms. - Recall that \(\ln(e^2)=2\).

1. The square root requires \(x\ge0\), and the logarithm’s argument must be strictly positive. Therefore, \(x>0\), so \(D_g=(0, \infty)\). 2. Use quotient and power properties: \(g(x)=\ln(\sqrt{x})-\ln(e^2)=\frac{1}{2}\ln x-2\). 3. Differentiate: \(g'(x)=\frac{1}{2x}\).

1. \(D_g=(0, \infty)\) 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. - Recall \(\ln(e^2)=2\).

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

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

- 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) \(f'(x)=\frac{3}{4}\sin x\)

- Recall the derivative of the sine function. - Evaluate the derivative at \(x=0\). - A line through the origin has a particularly simple y-intercept.

1. The point lies on the graph because \(f(0)=\sin(0)=0\). 2. Differentiate: \(f'(x)=\cos(x)\). 3. The tangent slope is \(f'(0)=\cos(0)=1\). 4. A line through the origin with slope \(1\) has equation \(y=x\).

\(y=x\)

- First differentiate both functions. - Use the slope and intercepts to identify the linear graph. - Use the value and direction near \(x = 0\) to distinguish \(-2\sin x\) from \(2\sin x\).

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 \(1.5\). This 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 of the origin. This is graph (2).

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

- Set the function equal to zero and isolate the exponential expression. - Apply the natural logarithm. - Differentiate the exponential function using the chain rule.

1. Set \(h(x)=0\): \(3-e^{x-1}=0\), so \(e^{x-1}=3\). 2. Taking natural logarithms gives \(x-1=\ln(3)\), so the \(x\)-intercept is \((1+\ln(3), 0)\). 3. Differentiate: \(h'(x)=-e^{x-1}\).

\(x\)-intercept: \((1+\ln(3), 0)\); \(h'(x)=-e^{x-1}\)

- 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, \(f'(x)=2e^{2x+5}\).

- Substitute a convenient x-value to check the proposed intersection. - What graph transformation is represented by replacing \(x\) with \(-x\)? - Compare the signs of the derivatives of the two functions. - Use exponent rules to simplify the product of the function expressions. - Distinguish between behavior as \(x\to\infty\) and as \(x\to-\infty\).

1. Since \(f(0)=e^0=1\) and \(g(0)=e^{-0}=1\), both graphs contain \((0, 1)\). Statement 1 is true. 2. Because \(g(x)=f(-x)\), the graph of \(g\) is the reflection of the graph of \(f\) across the y-axis, not the x-axis. Statement 2 is false. 3. The derivatives are \(f'(x)=e^x>0\) and \(g'(x)=-e^{-x}<0\) for every real \(x\). Their slopes can never be equal. Statement 3 is false. 4. Using exponent rules, \(h(x)=e^xe^{-x}=e^0=1\). The graph of \(h(x)=1\) is horizontal. Statement 4 is true. 5. As \(x\to\infty\), \(e^x\to\infty\), while \(e^{-x}\to0\). Only the graph of \(g\) approaches the x-axis. Statement 5 is false.

1. True 2. False 3. False 4. True 5. False

- Use the chain rule for a function of the form \(e^{kx}\). - How can you check whether a graph contains \((0, 0)\)? - What does the sign of the first derivative tell you about monotonicity? - What must happen to concavity at an inflection point? - How does a vertical shift affect the range?

1. Applying the chain rule gives \(g'(x)=e^{3x}\cdot3=3e^{3x}\). Statement a is true. 2. Since \(h(0)=e^0-1=0\), the graph contains \((0, 0)\). Statement b is true. 3. A function is strictly increasing on an interval when its derivative is positive throughout that interval. Since \(f'(x)=e^x>0\) for every real \(x\), statement c is true. 4. The second derivative is \(f''(x)=e^x>0\) for every real \(x\), so the graph is concave up everywhere and has no inflection point. Statement d is false. 5. The range of \(e^x\) is \((0,\infty)\). Shifting the graph down \(5\) units gives the range \((-5,\infty)\). Statement e is true.

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

- Use the chain rule for an exponential function with \(2x\) in the exponent. - Write the tangent equation at \(x=0\). - Negative slope gives a sign condition on \(k\). - Reverse an inequality when multiplying by a negative quantity. - Set the tangent equation equal to \(0\) to find its x-intercept.

1. Differentiate twice: \(f_k'(x)=2ke^{2x}\) and \(f_k''(x)=4ke^{2x}\). Therefore, \(f_k''(0)=4k\). 2. At \(x=0\), the point is \((0, k-4)\) and the slope is \(2k\). The tangent line is \(y=2kx+k-4\). 3. Negative slope requires \(k<0\). 4. The x-intercept is \(x=\frac{4-k}{2k}\). Require \(\frac{4-k}{2k}<-1\). 5. Because \(2k<0\), multiplying reverses the inequality: \(4-k>-2k\), so \(k>-4\). 6. Combining the conditions gives \(-4<k<0\).

a) \(f_k''(0)=4k\) b) \(-4<k<0\)

- Compare the outputs as the x-values increase. - For consecutive x-values, how is the ratio of the outputs related to the base? - Substitute one point after finding \(a\) to determine \(b\). - Use \(a=e^{\ln(a)}\) to rewrite the function. - Apply the chain rule to differentiate \(e^{kx}\).

1. The function value increases from \(45\) at \(x=1\) to \(135\) at \(x=2\), so the function represents exponential growth. 2. Increasing \(x\) by \(1\) multiplies the output by \(a\). Therefore, \(a=\frac{135}{45}=3\). 3. Substitute \((1, 45)\): \(45=b\cdot3\), so \(b=15\). Thus, \(f(x)=15\cdot3^x\). 4. Since \(3=e^{\ln(3)}\), \(f(x)=15e^{x\ln(3)}\). 5. Differentiate: \(f'(x)=15\ln(3)e^{x\ln(3)}\). Therefore, \(f'(0)=15\ln(3)\approx16.48\).

a) Exponential growth, because \(135>45\). b) \(a=3\) and \(b=15\) c) \(f(x)=15e^{x\ln(3)}\); \(f'(x)=15\ln(3)e^{x\ln(3)}\); \(f'(0)=15\ln(3)\approx16.48\)

- The first derivative gives the slope. - What is the maximum value of the cosine function? - Express the maximum slope as a function of \(k\).

1. Apply the chain rule: \(f_k'(x)=k\cos(kx)\). 2. Since the maximum value of cosine is \(1\), the maximum value of the derivative is \(k\). Therefore, the maximum slope is \(k\). 3. Replacing \(k\) by \(2k\) changes the maximum slope from \(k\) to \(2k\). 4. Thus, doubling \(k\) doubles the maximum slope.

The maximum slope is \(k\). When \(k\) is doubled, the maximum slope doubles to \(2k\).

- Set the two function values equal to find their intersection. - A substitution involving \(e^x\) can turn the intersection equation into a quadratic equation. - Compare both the function values and derivative values at the intersection. - Parallel lines have equal slopes.

1. Set the functions equal: \(e^x=2-e^{-x}\). Multiplying by \(e^x\) gives \(e^{2x}-2e^x+1=0\). Let \(u=e^x\). Then \((u-1)^2=0\), so \(u=1\), \(x=0\), and the intersection point is \(S(0, 1)\). 2. Differentiate: \(f'(x)=e^x\) and \(g'(x)=e^{-x}\). Thus \(f'(0)=g'(0)=1\), so the graphs share the tangent line \(y=x+1\) at \(S\). 3. To compare the graphs, \(f(x)-g(x)=e^x+e^{-x}-2\geq0\), with equality only at \(x=0\). Therefore, the graph of \(f\) stays above the graph of \(g\), and the graphs touch without crossing at \(S\). 4. At the same x-value, parallel tangent lines require \(f'(x)=g'(x)\). Solving \(e^x=e^{-x}\) gives \(e^{2x}=1\), so \(x=0\) is the only solution.

a) \(S(0, 1)\) b) \(f'(0)=g'(0)=1\). The graphs share the tangent line \(y=x+1\), and \(f(x)\geq g(x)\) with equality only at \(x=0\), so they do not cross. c) No. At the same x-value, the tangent lines are parallel only when \(x=0\).

- 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 \(y=\frac{e^{k+1}}{k}(x-k)\).

- Determine when the fraction inside the logarithm is positive. - Remember that \(x=0\) is excluded. - Use the quotient property of logarithms. - Rewrite \(\ln(x^2)\) as \(2\ln|x|\).

1. The logarithm’s argument must be positive. Since \(x^2>0\) for \(x\ne0\), the numerator must satisfy \(x+2>0\). Therefore, \(D_f=(-2, 0)\cup(0, \infty)\). 2. Rewrite using logarithm properties: \(f(x)=\ln(x+2)-2\ln|x|\). 3. Differentiate: \(f'(x)=\frac{1}{x+2}-\frac{2}{x}\). Combining the fractions gives \(f'(x)=-\frac{x+4}{x(x+2)}\).

\(D_f=(-2, 0)\cup(0, \infty)\) \(f'(x)=-\frac{x+4}{x(x+2)}\)

- 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. \(a\ln\!\left(1+\frac{p}{100}\right)\)

- What does replacing \(x\) with \(-x\) do to a graph? - Apply the chain rule to \(e^{-x}\). - Set the two function expressions equal and solve. - Use the first and second derivative tests to classify the critical point. - Apply the product rule for powers with the same base.

1. Since \(g(x)=e^{-x}=f(-x)\), the graph of \(g\) is the reflection of the graph of \(f\) across the y-axis. Statement 1 is true. 2. The derivatives are \(f'(x)=e^x=f(x)\) and \(g'(x)=-e^{-x}=-g(x)\). Statement 2 is true. 3. Solve \(e^x=e^{-x}\). This gives \(e^{2x}=1\), so \(2x=0\) and \(x=0\). Because the exponential function is one-to-one, this is the only solution. Statement 3 is true. 4. \(s'(x)=e^x-e^{-x}\), so \(s'(0)=0\). Also, \(s''(x)=e^x+e^{-x}>0\), so \(s\) has a local minimum, not a maximum, at \(x=0\). Statement 4 is false. 5. \(f(x)g(x)=e^xe^{-x}=e^0=1\). Statement 5 is true.

(1) True (2) True (3) True (4) False (5) True

- Use the sign of the first derivative to determine monotonicity. - Examine the end behavior as \(x\to-\infty\). - Evaluate the first derivative at the specified input. - Solve \(h(x)=0\) and use logarithm properties. - What does the sign of the second derivative tell you about concavity?

1. \(h'(x)=-0.5e^{0.5x}<0\) for every real \(x\), so \(h\) is strictly decreasing. Statement 1 is true. 2. As \(x\to-\infty\), \(e^{0.5x}\to0\), so \(h(x)\to2\). The horizontal asymptote is \(y=2\), not \(y=0\). Statement 2 is false. 3. \(h'(0)=-0.5e^0=-0.5\). Statement 3 is true. 4. Solving \(2-e^{0.5x}=0\) gives \(e^{0.5x}=2\), so \(0.5x=\ln(2)\) and \(x=2\ln(2)=\ln(4)\). Statement 4 is true. 5. \(h''(x)=-0.25e^{0.5x}<0\) for every real \(x\), so the graph is concave down. Statement 5 is true.

(1) True (2) False (3) True (4) True (5) True

- 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 \((1, 0)\).

- Set the first derivative equal to zero. - Use the sign of the second derivative to classify the critical point. - Replace the parameter with the maximum x-coordinate. - A common point must have a function value independent of \(a\).

1. Differentiate: \(f_a'(x)=\frac{a}{x}-1\). The critical-point equation gives \(x=a\). Since \(f_a''(x)=-\frac{a}{x^2}<0\), this point is a local maximum. 2. Its y-coordinate is \(a\ln a-a\), so the local maximum is \((a, a\ln a-a)\). Replacing \(a\) with the maximum x-coordinate gives the locus \(y=x\ln x-x\), where \(x>0\). 3. A common point must make the coefficient of \(a\) equal to zero. Thus \(\ln x=0\), so \(x=1\). Then \(f_a(1)=-1\), giving the common point \((1,-1)\).

a) \((a, a\ln a-a)\) b) \(y=x\ln x-x\) for \(x>0\) c) \((1,-1)\)

- Use the chain rule when differentiating \(e^{2x}\). - Solve the exponential equation by taking a logarithm. - Use the critical-point equation to simplify the y-coordinate. - An inflection point requires a change in concavity.

1. Differentiate: \(g_a'(x)=1-2ae^{2x}\) and \(g_a''(x)=-4ae^{2x}\). 2. The critical-point equation gives \(e^{2x}=\frac{1}{2a}\), so \(x=-\frac{1}{2}\ln(2a)\). Since \(g_a''(x)<0\) for all \(x\), this is the unique local maximum. 3. At the maximum, \(ae^{2x}=\frac{1}{2}\). Therefore, \(y=x-\frac{1}{2}\), so every maximum lies on \(y=x-\frac{1}{2}\). 4. Because \(g_a''(x)=-4ae^{2x}<0\) for every \(x\), the concavity never changes. Thus there are no inflection points.

a) \(\left(-\frac{1}{2}\ln(2a), -\frac{1}{2}\ln(2a)-\frac{1}{2}\right)\) b) \(y=x-\frac{1}{2}\) c) The second derivative is always negative, so the graphs have no inflection points.

- Set the first derivative equal to zero. - Use the one-to-one property of the exponential function. - Use the sign of the second derivative. - Replace \(\ln k\) using the minimum x-coordinate.

1. Differentiate: \(g_k'(x)=ke^x-1\). The critical-point equation gives \(e^x=\frac{1}{k}\), so \(x=-\ln k\). Because the exponential function is one-to-one, this is the only critical point. 2. Since \(g_k''(x)=ke^x>0\) for all \(x\), the critical point is a local minimum. 3. Its y-coordinate is \(g_k(-\ln k)=ke^{-\ln k}+\ln k=1+\ln k\). Since \(x=-\ln k\), this becomes \(y=1-x\).

a) The unique local minimum is \((-\ln k, 1+\ln k)\). b) \(y=1-x\)

- Rewrite the function so the coefficient and vertical shift are easy to identify. - What happens to the exponential term as time increases without bound? - Use the chain rule to differentiate the exponential expression. - In a time-dependent process, what does the first derivative represent?

1. Rewrite the function as \(U(t)=-15e^{-0.4t}+15\). Starting with \(g(t)=e^{-0.4t}\), reflect the graph across the \(t\)-axis, stretch it vertically by a factor of \(15\), and shift it up \(15\) units. 2. Since \(e^{-0.4t}\to 0\) as \(t\to\infty\), \(U(t)\to 15\). The horizontal asymptote is \(U=15\). It represents the limiting voltage of the capacitor. 3. Differentiate using the chain rule: \(U'(t)=15\left(0.4e^{-0.4t}\right)=6e^{-0.4t}\). 4. At \(t=2\), \(U'(2)=6e^{-0.8}\approx 2.70\,\text{V/s}\). At that instant, the voltage is increasing at about \(2.70\) volts per second.

a) Reflect across the \(t\)-axis, stretch vertically by a factor of \(15\), and shift up \(15\) units. b) The horizontal asymptote is \(U=15\). The capacitor voltage approaches \(15\,\text{V}\). c) \(U'(t)=6e^{-0.4t}\), and \(U'(2)\approx 2.70\,\text{V/s}\).

- 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.

- Use the derivative of \(\ln x\). - Apply the second derivative test. - Solve the minimum x-coordinate for \(t\). - Use logarithm properties to simplify the locus equation.

1. Differentiate: \(f_t'(x)=tx-\frac{1}{x}\). The critical-point equation gives \(tx^2=1\), so \(x=\frac{1}{\sqrt{t}}\). 2. Since \(f_t''(x)=t+\frac{1}{x^2}>0\), the critical point is a local minimum. 3. Its y-coordinate is \(\frac{1}{2}-\ln\left(t^{-1/2}\right)=\frac{1}{2}+\frac{1}{2}\ln t\). Thus the local minimum is \(\left(\frac{1}{\sqrt{t}}, \frac{1}{2}+\frac{1}{2}\ln t\right)\). 4. At the minimum, \(t=\frac{1}{x^2}\). Therefore, \(y=\frac{1}{2}+\frac{1}{2}\ln\left(\frac{1}{x^2}\right)=\frac{1}{2}-\ln x\), with \(x>0\).

a) \(\left(\frac{1}{\sqrt{t}}, \frac{1}{2}+\frac{1}{2}\ln t\right)\) b) \(y=\frac{1}{2}-\ln x\) for \(x>0\)

- Compare logarithmic growth with quadratic growth. - Use the derivative of \(\ln x\) and the domain restriction. - Apply the second derivative test. - Replace \(a\) using the maximum x-coordinate.

1. As \(x\to0^+\), \(\ln x\to-\infty\) while \(-\frac{1}{2}x^2\to0\), so \(f_a(x)\to-\infty\). As \(x\to\infty\), the negative quadratic term dominates the logarithmic term, so \(f_a(x)\to-\infty\). 2. Differentiate: \(f_a'(x)=\frac{a}{x}-x\). The critical-point equation gives \(x^2=a\), so \(x=\sqrt{a}\). 3. Since \(f_a''(x)=-\frac{a}{x^2}-1<0\), this point is a local maximum. Its y-coordinate is \(\frac{a}{2}\ln a-\frac{a}{2}\), giving the local maximum \(\left(\sqrt{a}, \frac{a}{2}(\ln a-1)\right)\). 4. At the maximum, \(a=x^2\). Therefore, \(y=\frac{x^2}{2}(\ln(x^2)-1)=x^2\ln x-\frac{1}{2}x^2\), with \(x>0\).

a) \(\lim_{x\to0^+}f_a(x)=-\infty\) and \(\lim_{x\to\infty}f_a(x)=-\infty\) b) \(\left(\sqrt{a}, \frac{a}{2}(\ln a-1)\right)\) c) \(y=x^2\ln x-\frac{1}{2}x^2\) for \(x>0\)

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

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

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 \(\frac{\pi}{2}\).

- 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 \(1\).

- Use the horizontal asymptote to identify \(c\). - Use the y-intercept to find \(a\). - Substitute the graph's zero to find \(b\). - Differentiate using the chain rule and evaluate at \(0\) and \(1\).

1. The graph approaches the horizontal asymptote \(y=4\) as \(x\to-\infty\), so \(c=4\). The y-intercept is \((0, 2)\), so \(a+4=2\) and \(a=-2\). 2. The graph has a zero at \(x=1\). Thus, \(-2e^b+4=0\), so \(e^b=2\) and \(b=\ln2\). 3. Therefore, \(f(x)=-2e^{(\ln2)x}+4\), and \(f'(x)=-2\ln2\,e^{(\ln2)x}\). Hence, \(f'(0)=-2\ln2\) and \(f'(1)=-4\ln2=2f'(0)\).

a) \(c=4\), \(a=-2\) b) \(b=\ln2\) c) \(f'(0)=-2\ln2\) and \(f'(1)=-4\ln2=2f'(0)\)

- Identify where the sine curve has horizontal tangents. - Estimate whether the curve is increasing or decreasing most steeply at the remaining points. - Plot the slope values and connect them with a smooth periodic curve.

1. The slopes are \(f^{\prime}(0)=1\), \(f^{\prime}\left(\frac{\pi}{2}\right)=0\), \(f^{\prime}(\pi)=-1\), \(f^{\prime}\left(\frac{3\pi}{2}\right)=0\), and \(f^{\prime}(2\pi)=1\). 2. Plotting the points \((0, 1)\), \(\left(\frac{\pi}{2}, 0\right)\), \((\pi, -1)\), \(\left(\frac{3\pi}{2}, 0\right)\), and \((2\pi, 1)\) and connecting them smoothly produces the cosine graph. 3. Therefore, \(f^{\prime}(x)=\cos(x)\).

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

- Set the first derivative equal to zero. - Use the natural logarithm to solve for \(x\). - Apply the second derivative test. - Replace \(a\) using the minimum x-coordinate.

1. Differentiate: \(g_a'(x)=e^x-a\). The critical-point equation gives \(e^x=a\), so \(x=\ln a\). 2. Since \(g_a''(x)=e^x>0\), this point is a local minimum. 3. Its y-coordinate is \(g_a(\ln a)=a-a\ln a\). Thus the local minimum is \((\ln a, a-a\ln a)\). 4. At the minimum, \(a=e^x\). Substitution gives \(y=e^x-e^x x=e^x(1-x)\).

a) \((\ln a, a-a\ln a)\) b) \(y=e^x(1-x)\)

- 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 \(\frac{1}{2}\).

- Evaluate the function at \(x=0\). - A change in an outside multiplier produces a vertical scaling. - Use the point and slope at \(x=0\) to write the tangent line.

1. Since \(g_k(0)=k\), the y-intercepts give p: \(k=2\) and q: \(k=4\). 2. Doubling \(k\) doubles every function value, which is a vertical stretch by a factor of \(2\). 3. The derivative is \(g_k'(x)=-ke^{-x}\). At \(x=0\), the point is \((0, k)\) and the slope is \(-k\). 4. The tangent line is \(y=-kx+k\). Setting \(y=0\) gives \(x=1\), independent of \(k\).

a) p: \(k=2\); q: \(k=4\) b) Vertical stretch by a factor of \(2\) c) The tangent is \(y=-kx+k\), with x-intercept \(1\).

- The horizontal asymptote identifies the vertical shift. - Use the y-intercept to find \(a\). - Substitute the given point and compare exponents. - Differentiate to find the tangent slope.

1. The horizontal asymptote is \(y=3\), so \(c=3\). The y-intercept is \((0, 1)\), so \(a+3=1\) and \(a=-2\). 2. Substitute \(P(2, 3-2e)\) into \(f(x)=-2e^{bx}+3\): \(3-2e=-2e^{2b}+3\). Thus, \(e=e^{2b}\), so \(b=\frac{1}{2}\). 3. The function is \(f(x)=-2e^{x/2}+3\), so \(f'(x)=-e^{x/2}\). At \(x=0\), the slope is \(-1\), and the point is \((0, 1)\). Therefore, the tangent line is \(y=-x+1\).

a) \(c=3\), \(a=-2\) b) \(b=\frac{1}{2}\) c) \(y=-x+1\)

- Use point-slope form for each tangent line. - Use symmetry about the \(y\)-axis. - Find the triangle area from its base and height. - The \(x\)-intercepts of the tangent lines determine the base.

1. Let \(c=\phi(1)=\phi(-1)=\frac{1}{\sqrt{2\pi e}}\). Since \(\phi'(x)=-x\phi(x)\), the slopes at \(-1\) and \(1\) are \(c\) and \(-c\), respectively. 2. a) At \((-1, c)\), the tangent line is \(y-c=c(x+1)\), so \(y=c(x+2)\). At \((1, c)\), the tangent line is \(y-c=-c(x-1)\), so \(y=-c(x-2)\). 3. b) The lines meet the \(x\)-axis at \((-2, 0)\) and \((2, 0)\), so the base has length \(4\). The tangent lines intersect at \((0, 2c)\), so the height is \(2c\). Therefore, \(A=\frac12\cdot4\cdot2c=4c=\frac{4}{\sqrt{2\pi e}}\approx0.968\).

a) \(y=\frac{1}{\sqrt{2\pi e}}(x+2)\) and \(y=-\frac{1}{\sqrt{2\pi e}}(x-2)\) b) \(A=\frac{4}{\sqrt{2\pi e}}\approx0.968\) square units

- Determine where sine and cosine have the same value. - Differentiate both functions to find the tangent slopes at the intersection. - The two slopes have equal magnitudes and opposite signs. - Use the formula for the acute angle between two lines.

1. Set the functions equal: \(\sin(x)=\cos(x)\). On \([0, \pi]\), this gives \(x=\frac{\pi}{4}\). 2. The y-coordinate is \(\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\). Thus, the intersection is \(\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)\). 3. The derivatives are \(f^{\prime}(x)=\cos(x)\) and \(g^{\prime}(x)=-\sin(x)\). At \(x=\frac{\pi}{4}\), the tangent slopes are \(m_f=\frac{\sqrt{2}}{2}\) and \(m_g=-\frac{\sqrt{2}}{2}\). 4. For the acute angle \(\alpha\) between two lines, \(\tan(\alpha)=\left|\frac{m_f-m_g}{1+m_fm_g}\right|\). Therefore, \(\tan(\alpha)=\frac{\sqrt{2}}{1-\frac{1}{2}}=2\sqrt{2}\). 5. Hence, \(\alpha=\arctan(2\sqrt{2})\approx70.5^\circ\).

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