Determine the following limits; show work or briefly explain your thinking on each one. If you apply L'Hopital's Rule, indicate where you have applied it and why you can apply it. If your final answer is “does not exist,” ∞, or, -∞, briefly explain your answer. (You will not receive full credit for a “does not exist” answer if the answer is ∞ or -∞.)
- \(\displaystyle \lim_{x \to 2} \sqrt[3]{x^2 - 6x - 7}\)
- \(\displaystyle \lim_{x \to 0} \frac{\sin(x) - x}{1 - \cos(x)}\)
- \(\displaystyle \lim_{x \to \infty} e^{-4x} + \ln(x)\)
- \(\displaystyle \lim_{x \to 0^+} \left( \frac{1}{x} + \frac{5}{x(x-5)} \right)\)
Solution:
- The cube-root function is continuous for every real input, so we may substitute \(x=2\) directly: \[ \lim_{x\to2}\sqrt[3]{x^2-6x-7} =\sqrt[3]{2^2-6(2)-7} =\sqrt[3]{4-12-7} =\sqrt[3]{-15}. \] Therefore the limit is \(\boxed{\sqrt[3]{-15}}\).
- As \(x\to0\), both numerator and denominator go to \(0\), so this is an indeterminate form. We may apply L'Hospital's Rule: \[ \lim_{x\to0}\frac{\sin x-x}{1-\cos x} =\lim_{x\to0}\frac{\cos x-1}{\sin x}. \] This is still \(0/0\), so apply L'Hospital's Rule again: \[ \lim_{x\to0}\frac{\cos x-1}{\sin x} =\lim_{x\to0}\frac{-\sin x}{\cos x} =0. \] Hence the limit is \(\boxed{0}\).
- As \(x\to\infty\), the exponential term \(e^{-4x}\) tends to \(0\), while the logarithm \(\ln x\) increases without bound. Therefore the sum behaves like \(\ln x\), so \[ \lim_{x\to\infty}\left(e^{-4x}+\ln x\right)=\boxed{+\infty}. \]
- First combine the two terms: \[ \frac1x+\frac{5}{x(x-5)} =\frac{x-5}{x(x-5)}+\frac{5}{x(x-5)} =\frac{x}{x(x-5)} =\frac{1}{x-5}, \] for \(x\neq0,5\). Therefore \[ \lim_{x\to0^+}\left(\frac1x+\frac{5}{x(x-5)}\right) =\lim_{x\to0^+}\frac{1}{x-5} =-\frac15. \] So the limit is \(\boxed{-\dfrac15}\).
Use the limit definition of the derivative to determine the derivative of
Your answer should be \(\displaystyle f'(x) = \frac{1}{2\sqrt{x + 1}}\). No points will be awarded for the application of differentiation rules (and L'Hopital's Rule is not allowed.)
Show all steps.**
The graph below is the graph of the derivative of a function \(\displaystyle f\) (that is, the graph shows \(\displaystyle f'\)). The domain of this function is all real numbers.
DERIVATIVE
[A graph is shown here.]
- For what values of \(\displaystyle x\) does the original function \(\displaystyle f\) have a local maximum? Why?
- For what values of \(\displaystyle x\) does the original function \(\displaystyle f\) have an inflection point? Why?
- For what values of \(\displaystyle x\) is the tangent line to the graph of \(\displaystyle f\) horizontal? Why?
IMAGE
Solution: To use the limit definition, begin with \[ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} =\lim_{h\to0}\frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}. \] Multiply numerator and denominator by the conjugate: \[ \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\cdot \frac{\sqrt{x+h+1}+\sqrt{x+1}}{\sqrt{x+h+1}+\sqrt{x+1}}. \] The numerator becomes \[ (x+h+1)-(x+1)=h, \] so for \(h\neq0\), \[ \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h} =\frac{1}{\sqrt{x+h+1}+\sqrt{x+1}}. \] Now let \(h\to0\): \[ f'(x)=\lim_{h\to0}\frac{1}{\sqrt{x+h+1}+\sqrt{x+1}} =\frac{1}{2\sqrt{x+1}}. \] Therefore \[ \boxed{f'(x)=\frac{1}{2\sqrt{x+1}}}. \] For the graph questions, we use the fact that the picture shown is the graph of \(f'\), not of \(f\). A local maximum of \(f\) occurs where \(f'\) changes sign from positive to negative. From the graph, that happens at approximately \(\boxed{x=2}\). An inflection point of \(f\) occurs where \(f''\) changes sign, which we detect on the graph of \(f'\) as a local maximum or local minimum of \(f'\). From the picture, these occur approximately at \(\boxed{x=-2,0,1,3}\). The tangent line to \(f\) is horizontal where \(f'(x)=0\). So we look for the \(x\)-intercepts of the graph of \(f'\). Those occur approximately at \(\boxed{x=-3,0,2}\).
Determine \(\displaystyle \frac{dy}{dx}\) for each equation below. Remember to use correct notation to write your final answer. You do not need to simplify your final answer.
- \(\displaystyle y = \frac{2}{x^5} - \sqrt[3]{x}\)
- \(\displaystyle y = \frac{e^{-5x}}{\sin(x^2)}\)
- \(\displaystyle y = \ln(x)\tan(x)\)
- \(\displaystyle y = \sqrt{\cos^2(x)+2x}\)
Determine \(\displaystyle \frac{dy}{dx}\) for each equation below. Remember to use correct notation to write your final answer. You do not need to simplify your final answer.
- \(\displaystyle x^3 + y^3 = 2x + x \ln(y)\)
- \(\displaystyle y = \int_{2}^{x} \left( \sqrt{e^t} + \frac{1}{1 + 4t^2} \right) dt\)
Solution:
- \(\displaystyle y'= -\frac{10}{x^6}-\frac{1}{3x^{2/3}}\).
- \(\displaystyle y'=\frac{e^{-5x}}{\sin(x^2)}\left(-5-2x\cot(x^2)\right)\).
- \(\displaystyle y'=\frac{\tan x}{x}+\ln(x)\sec^2 x\).
- \(\displaystyle y'=\frac{1-\sin x\cos x}{\sqrt{\cos^2 x+2x}}\).
- Implicitly: \(3x^2+3y^2y'=2+\ln y+\frac{x}{y}y'\), so \(\displaystyle y'=\frac{2+\ln y-3x^2}{3y^2-x/y}\).
- By the FTC, \(\displaystyle y'=\sqrt{e^x}+\frac{1}{1+4x^2}=e^{x/2}+\frac{1}{1+4x^2}\).
Evaluate the following. You do not need to simplify your answers.
Evaluate the following. You do not need to simplify your answers.
- \(\displaystyle \int \frac{3 \cos(\ln(x))}{x} dx\)
- \(\displaystyle \int_{-1}^{0} e^{-3x+1} dx\)
4. (4 points) The graph below shows the rate of stock price change for Gimble Company in dollars per month.
DERIVATIVE
VISUAL_REF: (graph)
- On what interval(s) is the stock price increasing? On what interval(s) is the stock price decreasing?
- At what time(s) is the stock price at a local maximum? At what time(s) is the stock price at a local minimum?
- Sketch a possible graph of the stock price.
Label axes and indicate at least one value of \(\displaystyle t\) for each local max/min identified above.
- State the meaning of a local maximum and a local minimum for the derivative in the context of the problem.
IMAGE
Solution:
- Integrate term by term: \[ \int \left(5+\frac{1}{3x^{1/5}}-\frac9x+x^2\right)dx =\int 5\,dx+\frac13\int x^{-1/5}\,dx-9\int\frac1x\,dx+\int x^2\,dx. \] This gives \[ 5x+\frac13\cdot\frac{x^{4/5}}{4/5}-9\ln|x|+\frac{x^3}{3}+C =\boxed{5x+\frac{5}{12}x^{4/5}-9\ln|x|+\frac{x^3}{3}+C}. \]
- Again integrate term by term: \[ \int\left(\sin x-\frac{1}{\sqrt{1-x^2}}\right)dx =\int\sin x\,dx-\int\frac{1}{\sqrt{1-x^2}}\,dx. \] Since \(\int\sin x\,dx=-\cos x\) and \(\int \frac{1}{\sqrt{1-x^2}}\,dx=\arcsin x\), we obtain \[ \boxed{-\cos x-\arcsin x+C}. \]
- Use the substitution \(u=\ln x\). Then \(du=\dfrac{dx}{x}\). So \[ \int \frac{3\cos(\ln x)}{x}\,dx =3\int \cos u\,du =3\sin u+C =\boxed{3\sin(\ln x)+C}. \]
- Antidifferentiate: \[ \int e^{-3x+1}\,dx=-\frac13 e^{-3x+1}+C, \] because the derivative of \(-3x+1\) is \(-3\). Therefore \[ \int_{-1}^{0}e^{-3x+1}\,dx =\left[-\frac13e^{-3x+1}\right]_{-1}^{0} =-\frac13 e^1-\left(-\frac13 e^4\right) =\frac{e^4-e}{3}. \] So the value is \(\boxed{\dfrac{e^4-e}{3}}\).
- The graph shown is the derivative of the stock price, so the stock price itself is increasing where the derivative is positive and decreasing where the derivative is negative. From the picture, the derivative is positive from about \(t=0\) until about \(t=6\), and negative after that. Therefore the stock price is increasing on approximately \(\boxed{(0,6)}\) and decreasing after about \(\boxed{t=6}\). A local maximum of the stock price occurs where its derivative changes from positive to negative, so that happens near \(\boxed{t=6}\). There is no interior local minimum visible on the interval shown. For the context question: a local maximum of the derivative means the stock price is increasing at the greatest rate near that time, while a local minimum of the derivative means the stock price is decreasing at the greatest rate near that time.
Each side of a square is increasing at a rate of 6 cm/sec. At what rate is the area of the square increasing when the area of the square is 16 cm\(\displaystyle ^2\)? Your final answer should be a complete sentence with appropriate units.
Solution: Let \(s\) be the side length of the square. Then the area is \[ A=s^2. \] Differentiate both sides with respect to time: \[ \frac{dA}{dt}=2s\frac{ds}{dt}. \] We are told that \(\dfrac{ds}{dt}=6\) cm/s. We are also told that the area is \(16\text{ cm}^2\), so \[ s^2=16 \quad\Longrightarrow\quad s=4, \] since side length must be positive. Now substitute into the related-rates formula: \[ \frac{dA}{dt}=2(4)(6)=48. \] Therefore, the area of the square is increasing at a rate of \(\boxed{48\text{ cm}^2/\text{s}}\).
A cylindrical-shaped tube with radius \(\displaystyle r\) and height \(\displaystyle h\) is made of a long rolled-up sheet plus two circular lids, held closed with tape. We need to put tape around the rims of both lids, as well as one long strip of tape down the side of the tube. (See the labels on the figure on the below.)
Tape on whole rim
Vertical tape strip
Tape on whole rim
- Given that we have \(\displaystyle 20\pi\) inches of tape to seal this tube shut, write a function for the VOLUME of the tube as a function of the radius \(\displaystyle r\).
- Determine an appropriate domain for your function from (a). Briefly explain your reasoning.
- NOTE: Your work from parts (a) and (b) will not be used in this question.
The amount of tape needed to create a \(\displaystyle 700\pi\) in\(\displaystyle ^3\) tube with radius \(\displaystyle r\) is given by
The domain of \(\displaystyle T(r)\) is \(\displaystyle (0,\infty)\). Determine the radius \(\displaystyle r\) of the cylinder that will minimize the amount of tape needed. \textit{Justify, using methods of calculus, why your dimensions yield the absolute minimum.}
IMAGE
Solution:
- The tube needs tape around both circular rims, plus one vertical strip down the side. Each rim has circumference \(2\pi r\), so the two rims require \(4\pi r\) inches of tape. The vertical strip has length \(h\). Thus the total tape constraint is \[ 4\pi r+h=20\pi. \] Solve for \(h\): \[ h=20\pi-4\pi r. \] The volume of the cylinder is \[ V=\pi r^2 h. \] Substitute the expression for \(h\): \[ V(r)=\pi r^2(20\pi-4\pi r)=20\pi^2r^2-4\pi^2r^3=4\pi^2r^2(5-r). \] Therefore \[ \boxed{V(r)=4\pi^2r^2(5-r)}. \]
- A physical cylinder must have positive radius and positive height. So we need \[ r>0 \quad\text{and}\quad h=20\pi-4\pi r>0. \] The second inequality gives \[ 20\pi-4\pi r>0 \quad\Longrightarrow\quad 5-r>0 \quad\Longrightarrow\quad r<5. \] Hence a reasonable domain is \[ \boxed{0<r<5}. \]
- We are given \[ T(r)=4\pi r+\frac{700}{r^2}, \qquad r>0. \] To minimize the amount of tape, differentiate: \[ T'(r)=4\pi-\frac{1400}{r^3}. \] Set the derivative equal to zero: \[ 4\pi-\frac{1400}{r^3}=0 \quad\Longrightarrow\quad 4\pi=\frac{1400}{r^3} \quad\Longrightarrow\quad r^3=\frac{350}{\pi}. \] So the critical point is \[ \boxed{r=\left(\frac{350}{\pi}\right)^{1/3}}. \] To justify that this gives an absolute minimum, compute the second derivative: \[ T''(r)=\frac{4200}{r^4}. \] For every \(r>0\), we have \(T''(r)>0\), so \(T\) is concave up on its entire domain. Therefore this critical point gives the absolute minimum amount of tape. The minimizing radius is \[ \boxed{r=\left(\frac{350}{\pi}\right)^{1/3}}. \]