If \(\displaystyle y = 2x^3 + 3x^2 - 12x - 10\), find \(\displaystyle y'(-1)\).
Solution: We differentiate: \(y'(x)=6x^2+6x-12\). Hence \(y'(-1)=6(1)+6(-1)-12=-12\). The correct choice is \(\boxed{-12}\).
If \(\displaystyle f(\theta) = \sin \theta\), find \(\displaystyle f^{(11)}(\pi)\), the 11th derivative of \(\displaystyle f(\theta)\) at \(\displaystyle \theta = \pi\).
Solution: The derivatives of \(\sin\theta\) cycle every four steps: \(\sin,\cos,-\sin,-\cos\). Since \(11\equiv 3\pmod 4\), the 11th derivative is \(-\cos\theta\). Therefore \(f^{(11)}(\pi)=-\cos\pi=-(-1)=\boxed{1}\).
How many critical points of \(\displaystyle y = x^5 - \frac{20}{3}x^3\) are NOT local extrema?
Solution: We compute \(y'(x)=5x^4-20x^2=5x^2(x^2-4)\). The critical points are \(x=-2,0,2\). At \(x=\pm2\), the sign of \(y'\) changes, so those are local extrema. At \(x=0\), the factor \(x^2\) does not change sign, so \(x=0\) is critical but not an extremum. Thus exactly \(\boxed{1}\) critical point is not a local extremum.
If \(\displaystyle y(x) = \sqrt{1-x^2} + x \arcsin{x}\), compute \(\displaystyle y'(x)\).
Solution: Differentiate term by term. First, \(\dfrac{d}{dx}\sqrt{1-x^2}=\dfrac{-x}{\sqrt{1-x^2}}\). Next, by the product rule, \(\dfrac{d}{dx}\bigl(x\arcsin x\bigr)=\arcsin x+\dfrac{x}{\sqrt{1-x^2}}\). The rational terms cancel, so \(y'(x)=\boxed{\arcsin x}\).
A person walks away from a lamp post at a rate of 3 ft/sec. The lamp post is 10 ft tall, and the person is 5 ft tall. How fast is the length of the person's shadow increasing when the person is 15 ft from the lamp post?
Solution: Let \(x\) be the distance from the lamp post to the person and \(s\) the shadow length. By similar triangles, \(\dfrac{10}{x+s}=\dfrac{5}{s}\), hence \(10s=5x+5s\), so \(s=x\). Differentiating gives \(\dfrac{ds}{dt}=\dfrac{dx}{dt}=3\) ft/sec. Therefore the shadow length is increasing at \(\boxed{3\text{ ft/sec}}\).
What is the slope of the tangent line to \(\displaystyle x y^2 + e^y = x^2\) at the point \(\displaystyle (1, 0)\)?
Solution: Differentiate implicitly: \(\dfrac{d}{dx}(xy^2)+\dfrac{d}{dx}(e^y)=\dfrac{d}{dx}(x^2)\). Thus \(y^2+2xyy'+e^y y'=2x\). At \((1,0)\), this becomes \(0+(0+1)y'=2\), so \(y'=\boxed{2}\).
Consider the function
What should \(\displaystyle a\) be so that the function is differentiable everywhere?
Solution: For differentiability at \(x=1\), we need continuity and matching one-sided derivatives. The left branch gives \(f(1)=1+a-3+3-a=1\), while the right branch gives \(\dfrac{2}{2}=1\), so continuity is automatic. The left derivative is \(2x+a-3\), hence at \(x=1\) it is \(a-1\). The right derivative of \(\dfrac{2x}{x^2+1}\) is \(\dfrac{2(1-x^2)}{(x^2+1)^2}\), which equals \(0\) at \(x=1\). Therefore \(a-1=0\), so \(\boxed{a=1}\).
If \(\displaystyle f(x) = x^{3} + 8x + \cos(3x)\) and \(\displaystyle g(x) = f^{-1}(x)\), find the slope of the tangent line to \(\displaystyle g(x)\) at the point \(\displaystyle (1, 0)\).
Solution: Since \(g=f^{-1}\), we use \(g'(y)=\dfrac{1}{f'(x)}\) where \(y=f(x)\). The point \((1,0)\) on \(g\) means \(g(1)=0\), so \(f(0)=1\), which is true. Now \(f'(x)=3x^2+8-3\sin(3x)\), hence \(f'(0)=8\). Therefore \(g'(1)=\dfrac{1}{8}\), so the slope is \(\boxed{\tfrac18}\).
Let \(\displaystyle f\) and \(\displaystyle g\) be differentiable functions, and suppose the following values are known:
Define \(\displaystyle h(x) = [ f(g(x)) ]^2\). What is the value of \(\displaystyle h'(a)\)?
Solution: We have \(h(x)=[f(g(x))]^2\). By the chain rule, \(h'(x)=2f(g(x))\,f'(g(x))\,g'(x)\). At \(x=a\), we know \(g(a)=c\), \(f(c)=3\), \(f'(c)=2\), and \(g'(a)=b\). Hence \(h'(a)=2\cdot3\cdot2\cdot b=\boxed{12b}\).
Below is the graph of the derivative \(\displaystyle f'\) of a continuous function \(\displaystyle f\).
Below is the graph of the derivative \(\displaystyle f'\) of a continuous function \(\displaystyle f\).
At what value(s) of \(\displaystyle x\) does \(\displaystyle f\) have a local maximum?
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Solution: A local maximum of \(f\) occurs where \(f'\) changes sign from positive to negative. From the graph, this happens at \(x=2\), where the curve crosses the axis from above to below, and also at \(x=6\), where the left-hand values of \(f'\) are positive and the right-hand values are negative. Thus the local maxima occur at \(\boxed{x=2\text{ and }x=6}\).
If \(\displaystyle H(x) = \int_{0}^{2x} e^{3t^2} dt\), find \(\displaystyle H'(1)\).
Solution: By the Fundamental Theorem of Calculus and the chain rule, \(H'(x)=e^{3(2x)^2}\cdot 2=2e^{12x^2}\). Evaluating at \(x=1\) gives \(H'(1)=\boxed{2e^{12}}\).
Let \(\displaystyle f(x)\) be a continuous function defined for all real numbers. We are given that
Calculate \(\displaystyle \int_{1}^{2} (3 - 2f(x)) dx\).
Solution: First compute \(\int_1^2 f(x)\,dx=\int_{-1}^2 f(x)\,dx-\int_{-1}^1 f(x)\,dx=5-3=2\). Then \(\int_1^2 (3-2f(x))\,dx=\int_1^2 3\,dx-2\int_1^2 f(x)\,dx=3-4=\boxed{-1}\).
Compute \(\displaystyle \int_{-2}^{0} 1 + \sqrt{4 - x^2} dx\) by recognizing it as the area of a region made up of simple geometric shapes.
Solution: On \([-2,0]\), the graph of \(y=\sqrt{4-x^2}\) is a quarter-circle of radius \(2\), so its area is \(\pi\). The term \(1\) contributes a rectangle of width \(2\) and height \(1\), hence area \(2\). Therefore \(\int_{-2}^{0}(1+\sqrt{4-x^2})dx=2+\pi=\boxed{\pi+2}\).
Compute the area bounded by \(\displaystyle y = e^x\), \(\displaystyle x = 0\), and \(\displaystyle y = 2\).
Solution: The region is bounded above by \(y=2\), below by \(y=e^x\), and on the left by \(x=0\). The curves meet when \(e^x=2\), i.e. at \(x=\ln 2\). Hence \(A=\int_0^{\ln 2}(2-e^x)\,dx=[2x-e^x]_0^{\ln 2}=2\ln 2-1\). So the area is \(\boxed{2\ln 2-1}\).
A scientist is studying how the concentration of a chemical in a solution changes with temperature. The concentration \(\displaystyle C\), measured in moles per liter, is modeled by the function
where \(\displaystyle T\) is the temperature in degrees Celsius. To make quick predictions without using a calculator, the scientist wants to use a linear approximation of \(\displaystyle C(T)\) near \(\displaystyle T = 2^\circ C\). Estimate the concentration at \(\displaystyle T = 2.1^\circ C\).
Solution: Let \(C(T)=\ln(T^2+5)\). Then \(C'(T)=\dfrac{2T}{T^2+5}\), so \(C'(2)=\dfrac49\). The linearization at \(T=2\) is \(L(T)=\ln 9+\dfrac49(T-2)\). At \(T=2.1\), we get \(C(2.1)\approx L(2.1)=\ln 9+\dfrac49(0.1)=\ln 9+\dfrac{2}{45}\). Therefore the estimate is \(\boxed{\ln 9+\tfrac{2}{45}}\).
\(\displaystyle \int_{0}^{2} e^{-x^2} dx\) is to be approximated using a Riemann sum with 8 subintervals of equal width, where the height of each rectangle is determined by the right endpoint of the subinterval. If \(\displaystyle i\) is an integer between 1 and 8, what is the area of the \(\displaystyle i\)th rectangle?
Solution: The interval \([0,2]\) is split into 8 equal parts, so \(\Delta x=\dfrac14\). The right endpoint of the \(i\)-th subinterval is \(x_i=\dfrac{i}{4}\). Therefore the area of the \(i\)-th rectangle is \(\Delta x\,e^{-x_i^2}=\dfrac14 e^{-(i/4)^2}\). Thus the answer is \(\boxed{\dfrac14 e^{-(i/4)^2}}\).
Evaluate \(\displaystyle \int_{0}^{1} \frac{x}{1 + x^2} dx\).
Solution: Use the substitution \(u=1+x^2\), so \(du=2x\,dx\) and \(x\,dx=\dfrac12du\). Then \(\int_0^1 \dfrac{x}{1+x^2}dx=\dfrac12\int_1^2 \dfrac{1}{u}du=\dfrac12[\ln u]_1^2=\boxed{\dfrac12\ln 2}\).
If \(\displaystyle f(x)\) is continuous and it is known that \(\displaystyle \int_{0}^{2} f(x) dx = 6\), evaluate \(\displaystyle \int_{0}^{\frac{\pi}{2}} f(2 \sin t) \cos t dt\).
Solution: Set \(u=2\sin t\). Then \(du=2\cos t\,dt\), so \(\cos t\,dt=\dfrac12du\). As \(t\) runs from \(0\) to \(\pi/2\), \(u\) runs from \(0\) to \(2\). Therefore \(\int_0^{\pi/2} f(2\sin t)\cos t\,dt=\dfrac12\int_0^2 f(u)\,du=3\). So the value is \(\boxed{3}\).
