Spring 24 Final Solution

Part I: Multiple Choice

Evaluate the following limits:

$\displaystyle \lim_{x \to 1} \frac{x^2 - 3}{2 - x}$

0 C -2 E DNE
1 D $\displaystyle \infty$ F Something else

$\displaystyle \lim_{x \to \infty} \frac{4x^2 - 10}{1 + 2x - 9x^2}$

$\displaystyle -\frac{4}{9}$ C 2 E $\displaystyle \infty$
$\displaystyle \frac{4}{9}$ D 4 F DNE

Solution:

This is a rational function, and the denominator is not zero at $x=1$ because $2-1=1$. Therefore direct substitution is valid: \[ \lim_{x\to1}\frac{x^2-3}{2-x} =\frac{1^2-3}{2-1} =\frac{-2}{1} =-2. \] So the limit is $\boxed{-2}$.
As $x\to\infty$, the highest-degree terms dominate. In the numerator the leading term is $4x^2$, and in the denominator the leading term is $-9x^2$. Therefore \[ \lim_{x\to\infty}\frac{4x^2-10}{1+2x-9x^2} =\frac{4}{-9} =-\frac49. \] Hence the limit is $\boxed{-\dfrac49}$.

Determine all horizontal asymptotes of $\displaystyle g(x) = \frac{\sqrt{9x^2+1}}{2x}$.

$\displaystyle y = -\frac{9}{2}$
$\displaystyle y = -\frac{3}{2}$
$\displaystyle y = \frac{3}{2}$
$\displaystyle y = \frac{9}{2}$
$\displaystyle y = -\frac{3}{2}$ and $\displaystyle y = \frac{3}{2}$
$\displaystyle y = -\frac{9}{2}$ and $\displaystyle y = \frac{9}{2}$
$\displaystyle y = 0$
No horizontal asymptotes

Solution: As $x\to+\infty$, $\sqrt{9x^2+1}\sim 3|x|=3x$, so $g(x)\to \dfrac{3x}{2x}=\dfrac32$. As $x\to-\infty$, $\sqrt{9x^2+1}\sim 3|x|=-3x$, so $g(x)\to \dfrac{-3x}{2x}=-\dfrac32$. Thus the horizontal asymptotes are $\boxed{y=-\tfrac32\text{ and }y=\tfrac32}$.

Below are the graphs of the derivatives of two functions $\displaystyle f(x)$ and $\displaystyle g(x)$. Carefully read the following statements and determine which of them must necessarily be true.

(Graph showing $\displaystyle g'(x)$ and $\displaystyle f'(x)$)

Statement (I): $\displaystyle f(x)$ has a local minimum.

Statement (II): $\displaystyle g(x)$ has a root.

Statement (III): $\displaystyle g(x)$ is increasing everywhere.

I only
II only
I and II
I and III
I, II, and III

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Two derivative graphs on coordinate axes

Solution: Statement (I) is true because the graph of $f'$ crosses the x-axis from negative to positive and later from positive to negative, so $f$ has a local minimum and a local maximum. Statement (II) need not be true: knowing only $g'$ does not determine whether $g$ itself crosses zero. Statement (III) is false because $g'$ is negative on part of the graph, so $g$ is not increasing everywhere. Therefore the correct choice is $\boxed{\text{I only}}$.

Determine $\displaystyle \frac{dy}{dx}$ if $\displaystyle x^4 = x^2 y - y^3$.

$\displaystyle \frac{dy}{dx} = \frac{2x}{1 - 3y^2}$
$\displaystyle \frac{dy}{dx} = \frac{4x^3}{2x - 3y^2}$
$\displaystyle \frac{dy}{dx} = \frac{4x^3 - 2xy}{x^2 - 3y^2}$
$\displaystyle \frac{dy}{dx} = \frac{4x^3 - x^2 y}{2x - 3y^2}$
$\displaystyle \frac{dy}{dx} = \frac{4x^3(x^2 - y^2) - x^4(2x - 2y)}{(x^2 - y^2)^2}$

Solution: Differentiate implicitly: $4x^3=2xy+x^2y'-3y^2y'$. Grouping the terms in $y'$, we obtain $(x^2-3y^2)y'=4x^3-2xy$. Hence $\displaystyle \frac{dy}{dx}=\boxed{\frac{4x^3-2xy}{x^2-3y^2}}$.

Consider the graph of a function $\displaystyle f(x)$ which is twice differentiable except at a few points on the interval $\displaystyle [-4, 5]$:

(Graph of $\displaystyle f(x)$ with axes and points at $\displaystyle x = -4, -1, 2, 5$)

Which of the following sketches best describes the graphs of $\displaystyle f'(x)$ and $\displaystyle f''(x)$?

IMAGE

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Set of candidate sketches for the first and second derivatives

Solution: On $(-4,-1)$, the graph is increasing and concave up, so $f'>0$ and increasing there. On $(-1,2)$, the graph is horizontal, so $f'=0$ and $f''=0$. On $(2,5)$, the graph is decreasing but flattening out, so $f'<0$ and increasing, hence $f''>0$. Among the sketches shown, this corresponds to the choice where $f'$ is positive on the left, zero in the middle, negative on the right, and $f''$ is positive on the curved pieces and zero on the flat piece. That is the option that matches the graph best.

At how many values of $\displaystyle x$ on the interval $\displaystyle [-1, 2]$ does the function $\displaystyle f(x) = \frac{1}{4} x^4 - x^3 + x^2 + 3$ attain absolute minimum?

$\displaystyle \circled{A}\ 0$

$\displaystyle \circled{B}\ 1$

$\displaystyle \circled{C}\ 2$

$\displaystyle \circled{D}\ 3$

$\displaystyle \circled{E}\ 4$

Solution: Compute $f'(x)=x^3-3x^2+2x=x(x-1)(x-2)$. On the interval $[-1,2]$, we must check $x=-1,0,1,2$. We get $f(-1)=\tfrac14+1+1+3=\tfrac{21}{4}$, $f(0)=3$, $f(1)=\tfrac14-1+1+3=\tfrac{13}{4}$, and $f(2)=4-8+4+3=3$. The absolute minimum value is $3$, attained at $x=0$ and $x=2$. Therefore the function attains its absolute minimum at $\boxed{2}$ values of $x$.

If $\displaystyle f(x) = \int_{\sqrt{x}}^{2} \sin(t^2) dt$ then $\displaystyle f'(x)$ is

$\displaystyle \circled{A}\ \frac{-\sin(x)}{\sqrt{x}}$

$\displaystyle \circled{B}\ \frac{-\sin(x)}{x}$

$\displaystyle \circled{C}\ \frac{\sin(x)}{x}$

$\displaystyle \circled{D}\ \frac{-\sin(x)}{2\sqrt{x}}$

$\displaystyle \circled{E}\ \frac{\sin(\sqrt{x})}{2\sqrt{x}}$

$\displaystyle \circled{F}\ \frac{2\sin(x)}{\sqrt{x}}$

$\displaystyle \circled{G}\ \frac{-\sin(x)}{2x}$

$\displaystyle \circled{H}\ \sin(x)$

$\displaystyle \circled{I}\ \sin(4) - \sin(x)$

Solution: By the Fundamental Theorem of Calculus with a variable lower limit, $f'(x)=-\sin((\sqrt{x})^2)\cdot \dfrac{d}{dx}(\sqrt{x})=-\sin(x)\cdot \dfrac{1}{2\sqrt{x}}$. Since $2\sqrt{x}=2x/\sqrt{x}$, this is also $-\dfrac{\sin x}{2\sqrt{x}}$. Thus $\boxed{f'(x)=-\dfrac{\sin(x)}{2\sqrt{x}}}$.

Approximate the area under the graph $\displaystyle f(x) = 1 + \sin^2(\pi x)$ on $\displaystyle \left[0, \frac{1}{2}\right]$ using a right-endpoint Riemann sum with three subintervals of equal length.

$\displaystyle \frac{1}{3}$ B $\displaystyle \frac{2}{3}$ C $\displaystyle \frac{5}{6}$ D $\displaystyle \frac{3}{4}$ E $\displaystyle \frac{2}{5}$

Solution: The interval $[0,\tfrac12]$ has length $\tfrac12$, so with three equal subintervals we have $\Delta x=\tfrac16$. The right endpoints are $\tfrac16,\tfrac13,\tfrac12$. Then $f(\tfrac16)=1+\sin^2(\pi/6)=1+\tfrac14=\tfrac54$, $f(\tfrac13)=1+\sin^2(\pi/3)=1+\tfrac34=\tfrac74$, and $f(\tfrac12)=1+\sin^2(\pi/2)=2$. Therefore the right-endpoint Riemann sum is $\tfrac16\left(\tfrac54+\tfrac74+2\right)=\tfrac16\cdot 5=\boxed{\tfrac56}$.

The rate of growth of rabbit population in a forest is given by $\displaystyle G(t) = 1000e^{0.05t}$, where $\displaystyle G(t)$ is measured in number of rabbits per year, and $\displaystyle t$ is the time in years since the start of 2010.

Using the correct units, what does $\displaystyle \int_2^7 G(t) dt$ represent?

The integral represents the area under the curve of $\displaystyle G(t)$ between the years 2012 and 2017, measured in square rabbits.
The total increase in the number of rabbits between the years 2012 and 2017, measured in rabbits.
The rate of growth of the rabbit population at year 2017, measured in rabbits per year.
The total number of rabbits born since 2012, measured in rabbits.
The average rate of growth of the rabbit population between the years 2012 and 2017, measured in rabbits per year.

Solution: Since $G(t)$ is measured in rabbits per year, the integral $\int_2^7 G(t)\,dt$ accumulates this rate over time. Therefore it represents the total increase in the rabbit population from year 2 to year 7 after 2010, i.e. from 2012 to 2017, measured in rabbits. So the correct interpretation is $\boxed{\text{the total increase in the number of rabbits between 2012 and 2017}}$.

Which of the following is the derivative of the function

\[\displaystyle f(x) = \frac{e^{\arctan(x)}}{1 + x^2}\]?

$\displaystyle f'(x) = \frac{e^{\arctan(x)}}{2x}$
$\displaystyle f'(x) = \frac{e^{\arctan(x)} (1 - 2x)}{(1 + x^2)^2}$
$\displaystyle f'(x) = \frac{e^{\arctan(x)}}{(1 + x^2)^2} \left( 1 - \frac{2x}{(1 + x^2)^2} \right)$
$\displaystyle f'(x) = e^{\arctan(x)} \left( \frac{1}{1 + x^2} - 2x \right)$
$\displaystyle f'(x) = \frac{e^{\arctan(x)} (1 + x^2 - 2x)}{(1 + x^2)^2}$

Solution: Write $f(x)=e^{\arctan x}(1+x^2)^{-1}$. Differentiate using the product rule: $\dfrac{d}{dx}e^{\arctan x}=\dfrac{e^{\arctan x}}{1+x^2}$, and $\dfrac{d}{dx}(1+x^2)^{-1}=-\dfrac{2x}{(1+x^2)^2}$. Hence $f'(x)=\dfrac{e^{\arctan x}}{(1+x^2)^2}-\dfrac{2xe^{\arctan x}}{(1+x^2)^2}=\boxed{\dfrac{e^{\arctan x}(1-2x)}{(1+x^2)^2}}$.

Which of the following is true about the function

\[\displaystyle H(x) = \frac{x^2 - 5x + 6}{x^2 - 8x + 15}?\]

$\displaystyle H(x)$ has a vertical asymptote at $\displaystyle x = 3$, but has a removable discontinuity at $\displaystyle x = 5$.
$\displaystyle H(x)$ has vertical asymptotes at $\displaystyle x = 3$ and $\displaystyle x = 5$, but has a removable discontinuity at $\displaystyle x = 2$.
$\displaystyle H(x)$ has a vertical asymptote at $\displaystyle x = 5$, but has a removable discontinuity at $\displaystyle x = 3$.
$\displaystyle H(x)$ has vertical asymptotes at $\displaystyle x = 2$ and $\displaystyle x = 3$, but has a removable discontinuity at $\displaystyle x = 5$.
$\displaystyle H(x)$ has a vertical asymptote at $\displaystyle x = 2$ and $\displaystyle x = 5$, but has a removable discontinuity at $\displaystyle x = 3$.

Solution: Factor numerator and denominator: $x^2-5x+6=(x-2)(x-3)$ and $x^2-8x+15=(x-3)(x-5)$. Thus $H(x)=\dfrac{x-2}{x-5}$ for $x\neq 3,5$. The factor $x-3$ cancels, so $x=3$ is a removable discontinuity. The denominator still vanishes at $x=5$, so $x=5$ is a vertical asymptote. Therefore the true statement is $\boxed{\text{vertical asymptote at }x=5\text{ and removable discontinuity at }x=3}$.

Classify all local extrema of $\displaystyle \frac{[\ln(x)]^2}{x}$.

$\displaystyle x = 1$ is a local minimum and $\displaystyle x = e^2$ is a local maximum
$\displaystyle x = 1$ is a local minimum and $\displaystyle x = 0$ and $\displaystyle x = e^2$ are local maxima
$\displaystyle x = 1$ is a local maximum and $\displaystyle x = e^2$ is a local minimum
$\displaystyle x = 1$ is the only extrema and is a local minimum
$\displaystyle x = e^2$ is the only local extrema and is a local maximum

Solution: Let $f(x)=\dfrac{(\ln x)^2}{x}$ for $x>0$. Then $f'(x)=\dfrac{2\ln x-(\ln x)^2}{x^2}=\dfrac{\ln x(2-\ln x)}{x^2}$. The critical points occur when $\ln x=0$ or $2-\ln x=0$, that is at $x=1$ and $x=e^2$. The sign of $f'$ is negative on $(0,1)$, positive on $(1,e^2)$, and negative on $(e^2,\infty)$. Hence $x=1$ is a local minimum and $x=e^2$ is a local maximum.

Let $\displaystyle F$ and $\displaystyle G$ be two differentiable functions satisfying

\[\displaystyle F\left(\frac{\pi}{4}\right) = -1, F'\left(\frac{\pi}{4}\right) = 3, G\left(\frac{\sqrt{2}}{2}\right) = 2, \text{ and } G'\left(\frac{\sqrt{2}}{2}\right) = 5.\]

Which of the following is the value of the derivative of $\displaystyle F(x) \cdot G(\cos(x))$ at $\displaystyle x = \pi/4$?

$\displaystyle 6 + \frac{5}{\sqrt{2}}$
$\displaystyle 6 - \frac{5}{\sqrt{2}}$
$\displaystyle 5 + \frac{6}{\sqrt{2}}$
$\displaystyle 5 - \frac{6}{\sqrt{2}}$
None of the above

Solution: Differentiate $F(x)G(\cos x)$ using the product and chain rules: $(FG\circ\cos)'=F'(x)G(\cos x)+F(x)G'(\cos x)(-\sin x)$. At $x=\pi/4$, this gives $3\cdot 2+(-1)\cdot 5\cdot\left(-\dfrac{\sqrt2}{2}\right)=6+\dfrac{5}{\sqrt2}$. Therefore the value is $\boxed{6+\dfrac{5}{\sqrt2}}$.

Find the area enclosed by the curve $\displaystyle y = 1 - x^2$, the x-axis, and the lines $\displaystyle x = 0$, and $\displaystyle x = 3$.

6 B -6 C 7 D $\displaystyle \frac{22}{3}$ E $\displaystyle \frac{16}{3}$

Solution: On $[0,1]$, the curve $y=1-x^2$ lies above the x-axis, and on $[1,3]$ it lies below. Thus the enclosed area is $\int_0^1(1-x^2)\,dx+\int_1^3(x^2-1)\,dx$. We compute $\int_0^1(1-x^2)\,dx=\left[x-\dfrac{x^3}{3}\right]_0^1=\dfrac23$, and $\int_1^3(x^2-1)\,dx=\left[\dfrac{x^3}{3}-x\right]_1^3=\dfrac{20}{3}$. Hence the total area is $\boxed{\dfrac{22}{3}}$.

Part II: Free Response

Compute the following limits:

(a) \[\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{x \sin(2x)}{\cos(x)}\]

(b) \[\displaystyle \lim_{x \to 0} x \sin\left(\frac{1}{x}\right)\]

Solution:

Use the double-angle identity $\sin(2x)=2\sin x\cos x$. Then \[ \frac{x\sin(2x)}{\cos x} =\frac{x(2\sin x\cos x)}{\cos x} =2x\sin x, \] for $x$ near $\pi/2$, where $\cos x\neq0$. Now both $x$ and $\sin x$ are continuous, so we can substitute $x=\pi/2$: \[ \lim_{x\to\pi/2}2x\sin x =2\cdot\frac{\pi}{2}\cdot1 =\pi. \] Therefore the limit is $\boxed{\pi}$.
For all $x\neq0$, we know \[ -1\le \sin\left(\frac1x\right)\le 1. \] Multiply through by $|x|\ge0$: \[ -|x|\le x\sin\left(\frac1x\right)\le |x|. \] As $x\to0$, both $-|x|$ and $|x|$ tend to $0$. By the Squeeze Theorem, \[ \lim_{x\to0}x\sin\left(\frac1x\right)=\boxed{0}. \]

Use the limit definition of the derivative to compute the derivative of $\displaystyle f(x) = x^3$ at $\displaystyle x = 2$. **No credit will be awarded for any other method.**

[Hint: recall that $\displaystyle (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$]*

Solution: Since $f(x)=x^3$, the derivative at $x=2$ is, by definition, \[ f'(2)=\lim_{h\to0}\frac{f(2+h)-f(2)}{h} =\lim_{h\to0}\frac{(2+h)^3-2^3}{h}. \] Now expand $(2+h)^3$: \[ (2+h)^3=8+12h+6h^2+h^3. \] Substitute this into the difference quotient: \[ \frac{(2+h)^3-8}{h} =\frac{8+12h+6h^2+h^3-8}{h} =\frac{12h+6h^2+h^3}{h}. \] For $h\neq0$, simplify: \[ \frac{12h+6h^2+h^3}{h}=12+6h+h^2. \] Now let $h\to0$: \[ f'(2)=\lim_{h\to0}(12+6h+h^2)=12. \] Therefore \[ \boxed{f'(2)=12}. \]

Consider the function $\displaystyle \ln(x)$.

Find its linearization at $\displaystyle x = 1$.
Use the above linearization to estimate $\displaystyle \ln(1.02)$.

Solution:

The linearization of a function $f$ at $x=a$ is \[ L(x)=f(a)+f'(a)(x-a). \] Here $f(x)=\ln x$ and $a=1$. Compute the needed values: \[ f(1)=\ln 1=0, \qquad f'(x)=\frac1x, \qquad f'(1)=1. \] Therefore \[ L(x)=0+1(x-1)=x-1. \] So the linearization is $\boxed{L(x)=x-1}$.
To estimate $\ln(1.02)$, use the linearization from part (a): \[ \ln(1.02)\approx L(1.02)=1.02-1=0.02. \] Thus \[ \boxed{\ln(1.02)\approx0.02}. \]

Evaluate the following indefinite integrals:

(a) \[\displaystyle \int \frac{x+1}{2x} dx\]

(b) \[\displaystyle \int \sqrt{2x+4} dx\]

Solution:

$\displaystyle \int \frac{x+1}{2x}\,dx=\frac12\int\left(1+\frac1x\right)dx=\boxed{\frac{x}{2}+\frac12\ln|x|+C}$.
Let $u=2x+4$, so $du=2dx$ and $dx=\dfrac12du$. Then $\int\sqrt{2x+4}\,dx=\dfrac12\int u^{1/2}du=\dfrac13 u^{3/2}+C=\boxed{\frac13(2x+4)^{3/2}+C}$.

In a petri dish, a colony of bacteria takes on a circular shape. The radius of this colony is increasing steadily at a rate of 0.2 millimeters per second. What’s the rate at which the area of the circle is increasing at the instant when the circumference reaches $\displaystyle 20\pi$ millimeters? Make sure to include the units in your final answer.

Solution: Let $r$ be the radius of the bacterial colony. Its area is \[ A=\pi r^2. \] Differentiate with respect to time: \[ \frac{dA}{dt}=2\pi r\frac{dr}{dt}. \] We are told that $\dfrac{dr}{dt}=0.2$ mm/s. We are also told that the circumference is $20\pi$ mm. Since circumference is \[ 2\pi r, \] we solve \[ 2\pi r=20\pi \quad\Longrightarrow\quad r=10\text{ mm}. \] Now substitute into the area-rate formula: \[ \frac{dA}{dt}=2\pi(10)(0.2)=4\pi. \] Therefore the area is increasing at a rate of $\boxed{4\pi\text{ mm}^2/\text{s}}$.

Determine the derivative of $\displaystyle f(x) = [\cos(2x)]^{3x}$. \textit{[You might want to use logarithmic differentiation.]}

Solution: Let $y=[\cos(2x)]^{3x}$. Take logarithms: $\ln y=3x\ln(\cos 2x)$. Differentiate: $\dfrac{y'}{y}=3\ln(\cos 2x)+3x\cdot\dfrac{-2\sin 2x}{\cos 2x}=3\ln(\cos 2x)-6x\tan(2x)$. Therefore $\boxed{y'=[\cos(2x)]^{3x}\bigl(3\ln(\cos 2x)-6x\tan(2x)\bigr)}$.

You have been put in charge of designing a box for a revolutionary product in your favorite company. The product manager tells you that the rectangular box needs to have volume 200 cm$\displaystyle ^3$, it should have a square base, and it should be painted in such a way that:

The top and bottom faces are squares, painted blue. (blue paint costs $1/cm$\displaystyle ^2$)

The front face is painted red (red paint costs $2/cm$\displaystyle ^2$), while the rear face is painted yellow (yellow paint costs $0.5/cm$\displaystyle ^2$).

The left and the right faces are not painted.

The aim is to minimize the cost of painting the box.

Draw a picture of the box. Mention explicitly which variable you use to denote which side, especially which variable you use to denote a side of the square base.
Write all the constraints and equations relevant to the situation.
Write a formula of the function you want to optimize in terms of a single variable, and give a reasonable domain for the function (this is the domain where the problem makes sense).
Minimize the function using calculus, and thus determine the dimensions of the box that minimize the cost of painting.

Solution:

Let $x$ be the side length of the square base, and let $h$ be the height of the box. Then the box has dimensions $x\times x\times h$.
The volume constraint is \[ x^2h=200. \] Because a physical box must have positive dimensions, we also require \[ x>0,\qquad h>0. \]
We now write the cost in terms of a single variable. The top and bottom are two square faces, each of area $x^2$, and blue paint costs \$1/cm$^2$. So together they cost \[ 2x^2. \] The front face has area $xh$ and red paint costs \$2/cm$^2$, so that face costs \[ 2xh. \] The rear face also has area $xh$, and yellow paint costs \$0.5/cm$^2$, so that face costs \[ 0.5xh. \] The left and right faces are not painted, so they contribute no cost. Therefore the total cost is \[ C=2x^2+2xh+0.5xh=2x^2+2.5xh. \] Use the volume constraint $h=\dfrac{200}{x^2}$: \[ C(x)=2x^2+2.5x\left(\frac{200}{x^2}\right) =2x^2+\frac{500}{x}. \] Thus the function to minimize is \[ \boxed{C(x)=2x^2+\frac{500}{x}}, \qquad \boxed{x>0}. \]
Differentiate: \[ C'(x)=4x-\frac{500}{x^2}. \] Set the derivative equal to zero: \[ 4x-\frac{500}{x^2}=0. \] Multiply by $x^2$: \[ 4x^3-500=0 \quad\Longrightarrow\quad 4x^3=500 \quad\Longrightarrow\quad x^3=125. \] So \[ x=5. \] Now find the corresponding height: \[ h=\frac{200}{x^2}=\frac{200}{25}=8. \] To justify that this critical point gives a minimum, compute the second derivative: \[ C''(x)=4+\frac{1000}{x^3}. \] Since $x>0$, we have $C''(x)>0$ for every $x$ in the domain. Therefore $C$ is concave up on $(0,\infty)$, so the critical point at $x=5$ gives the minimum cost. The dimensions that minimize the painting cost are \[ \boxed{x=5\text{ cm},\quad h=8\text{ cm}}. \]

Consider the function

\[\displaystyle f(x) = \frac{x^4}{2} + 2x^3.\]

Here is some useful information: the roots of $\displaystyle f(x)$ are $\displaystyle -4$ and $\displaystyle 0$, its domain is the set of all real numbers and its derivative is

\[\displaystyle f'(x) = 2x^3 + 6x^2.\]

Find the critical points of $\displaystyle f(x)$.
Find $\displaystyle f''(x)$ and determine where the curve is concave down.
Complete a number line/table/chart, partitioning it appropriately and indicating the sign of $\displaystyle f'$, the sign of $\displaystyle f''$ and the shape of $\displaystyle f$ (increasing/decreasing and concave up/down).
Sketch the graph of $\displaystyle f(x)$.

(Graph of coordinate axes with x and y labeled.)

IMAGE

Solution:

Critical points occur where $f'(x)=0$ or where $f'(x)$ is undefined. Since \[ f'(x)=2x^3+6x^2=2x^2(x+3), \] the derivative is defined for all real $x$. So we solve \[ 2x^2(x+3)=0. \] This gives \[ x=0 \quad\text{or}\quad x=-3. \] Therefore the critical points occur at $\boxed{x=-3\text{ and }x=0}$.
Differentiate again: \[ f''(x)=6x^2+12x=6x(x+2). \] The curve is concave down where $f''(x)<0$. Since $6>0$, we only need the sign of $x(x+2)$. This product is negative on the interval $(-2,0)$. Therefore $f$ is concave down on $\boxed{(-2,0)}$.
For monotonicity, study the sign of \[ f'(x)=2x^2(x+3). \] Because $2x^2\ge0$, the sign is controlled by $x+3$, except that $f'(0)=0$. Thus: \[ f'(x)<0 \text{ on } (-\infty,-3),\qquad f'(x)>0 \text{ on } (-3,0)\text{ and }(0,\infty). \] So $f$ decreases on $(-\infty,-3)$ and increases on $(-3,\infty)$. For concavity, study \[ f''(x)=6x(x+2). \] This is positive on $(-\infty,-2)$, negative on $(-2,0)$, and positive on $(0,\infty)$. Therefore $f$ is concave up on $(-\infty,-2)\cup(0,\infty)$ and concave down on $(-2,0)$.
The graph should reflect all the information found above: it crosses the $x$-axis at $-4$ and $0$, has a local minimum at $x=-3$, has inflection points where concavity changes, namely at $x=-2$ and $x=0$, decreases for $x<-3$, and increases for $x>-3$. That is the qualitative sketch required.