PROBLEM SOLVING STRATEGY: Maxima & Minima

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This is not a recipe for you simply to follow, but rather a series of steps to help guide your thinking that will often work well.

- Compute the function’s derivative, $f'(x)$.
- Find all critical numbers
*c*such that either:Make a list of the values- the derivative is zero: $f'(c) = 0$;
- or the derivative $f'(c)$ does not exist.

*c*that you find.

(If you are examining a closed interval ($a \le x \le b$), then you may be able to skip to step 4.) - For each critical point
*c,*check whether $f$ is a local maximum or minimum or neither. Use either the First Derivative Test or the Second Derivative Test.

**A. First Derivative Test**:

Check a convenient value that’s less than *c* to see whether $f’$ is positive or negative there; then check a convenient value that’s a greater than *c* to see if $f’$ is positive or negative there. As the figure below illustrates:

- if the derivative switches from positive to negative, you’ve found a local maximum.
- if it switches from negative to positive, you’ve found a local minimum.
- if it doesn’t switch, it is neither a maximum nor a minimum; it is simply a point with a horizontal tangent (if $f'(c) = 0$), or vertical tangent (if $f'(c)$ is undefined).

**B. Second Derivative Test**:

Compute the function’s second derivative, $f^″(x).$ Then determine the sign of the second derivative at *c,* $f^{″}(c)$. As the figure below illustrates:

- if $f^{″}(c)$ is negative (so the function is concave down there), then $c$ is a local maximum.
- if it’s positive (so the function is concave up there), then $c$ is a local minimum.

- If requested, compute the value(s) of $f(c)$ to find the $y$-values of the maxima and/or minima.
- If the function has endpoints ($a \le x \le b$), compute the values of $f(a)$ and $f(b)$ to see how they compare to the values of $f(c)$ you found step 4. The
**global**(or**absolute**) maximum or minimum may lie at one of these endpoints.

Question 1 steps you through the process to find the relative maximum and minimum of a function.

Question 2 examines the same function, but now looks for the global maximum and minimum on an interval.

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Question 1: *f(x)* = 2*x*^3 - 3*x*^2 -12*x* +1

Consider the function $f(x) = 2x^3 -3x^2 -12x +1$.**(a)** Find the critical points of $f$.**(b)** Use the First Derivative Test to determine whether each critical point is a local maximum, minimum, or neither.**(c)** Use the Second Derivative Test to verify your answer to part (b).**(a)** $ -1, \, 2$**(b)** $x = -1$: local maximum; $x = 2$: local minimum**(c)** See detailed solution.

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Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail

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To find the critical points of $f$, we must examine its derivative:
\begin{align*}
f'(x) &= \frac{d}{dx} \left[2x^3 -3x^2 -12x +1 \right] \\ \\
&= (2)(3)x^2 -3(2)x -12 \\ \\
&= 6x^2 -6x -12
\end{align*}
The critical points occur when $f'(c) = 0$. (Critical points could also occur where $f'(c)$ is undefined, but this polynomial is defined everywhere so we only have to consider when $f'(c) = 0$.)

\begin{align*} f'(c) = 6c^2 -6c -12 &= 0 \\ 6\left(x^2 – c -2 \right) &= 0\\ 6 (c -2)(c+1) &= 0\\ c &= 2, \, -1 \quad \cmark \end{align*}

\begin{align*} f'(c) = 6c^2 -6c -12 &= 0 \\ 6\left(x^2 – c -2 \right) &= 0\\ 6 (c -2)(c+1) &= 0\\ c &= 2, \, -1 \quad \cmark \end{align*}

Most people using the First Derivative Test find it helpful to create a small diagram like the one below to organize their thinking and results. We place the values of $x$ underneath the number line, marking the critical numbers we found earlier. Above the line we show information about $f'(x)$, starting with the fact that at these critical points $f’$ is zero.

The figure nicely illustrates that we need to consider the sign of $f’$ in three regions: (I) to the left of -1; (II) between -1 and 2; and (III) to the right of 2.

**(I)** An easy value to try to the left of -1 is $x =-2$:

\begin{align*} f'(-2) &= 6(-4)(-1) \\[8px] &> 0 \end{align*} We actually don’t care what the value of $f'(-2)$ is; we only care whether it’s positive or negative. Since it’s positive, we put a “+” above that region on our line:

We thus know that the function has a positive derivative, and so is increasing, on the interval $(-\infty, -1)$.

**(II)** A convenient value between -1 and 2 is zero:

\begin{align*} f'(0) &= 6(-2)(1) \\[8px] &< 0 \end{align*} Again, we don't care about the value specifically; without any further work we see that it's negative, and so we put a "–" above that region.

We thus know that the function has a negative derivative, and so is decreasing, on the interval $(-1, 2)$.

Furthermore, we now know that the critical point $x = -1$ is a relative maximum, according to the First Derivative Test, because $f’$ switches from positive to negative there.

**(III)** A convenient value to try to the right of 2 is $x=3$:

\begin{align*} f'(3) &= 6(1)(4) \\[8px] &>0 \end{align*} This value is positive, and so we put a “+” over that region on the line.

We now know that, after $x = 2$, the function always increases: it is increasing on $(2, \infty)$.

Furthermore, we conclude from the First Derivative Test that $x = 2$ is a relative minimum, since $f’$ switches from negative to positive there.

Summarizing our results, we conclude our First Derivative Tests:

$\bullet$ Since $f’$ switches from positive to negative at $x = -1$, this point is a local maximum. $\quad \cmark$

$\bullet$ Since $f’$ switches from negative to positive at $x = 2$, this point is a local minimum. $\quad \cmark$

For comparison, the figure below shows the graph of the function.

The figure nicely illustrates that we need to consider the sign of $f’$ in three regions: (I) to the left of -1; (II) between -1 and 2; and (III) to the right of 2.

\begin{align*} f'(-2) &= 6(-4)(-1) \\[8px] &> 0 \end{align*} We actually don’t care what the value of $f'(-2)$ is; we only care whether it’s positive or negative. Since it’s positive, we put a “+” above that region on our line:

We thus know that the function has a positive derivative, and so is increasing, on the interval $(-\infty, -1)$.

\begin{align*} f'(0) &= 6(-2)(1) \\[8px] &< 0 \end{align*} Again, we don't care about the value specifically; without any further work we see that it's negative, and so we put a "–" above that region.

We thus know that the function has a negative derivative, and so is decreasing, on the interval $(-1, 2)$.

Furthermore, we now know that the critical point $x = -1$ is a relative maximum, according to the First Derivative Test, because $f’$ switches from positive to negative there.

\begin{align*} f'(3) &= 6(1)(4) \\[8px] &>0 \end{align*} This value is positive, and so we put a “+” over that region on the line.

We now know that, after $x = 2$, the function always increases: it is increasing on $(2, \infty)$.

Furthermore, we conclude from the First Derivative Test that $x = 2$ is a relative minimum, since $f’$ switches from negative to positive there.

Summarizing our results, we conclude our First Derivative Tests:

$\bullet$ Since $f’$ switches from positive to negative at $x = -1$, this point is a local maximum. $\quad \cmark$

$\bullet$ Since $f’$ switches from negative to positive at $x = 2$, this point is a local minimum. $\quad \cmark$

For comparison, the figure below shows the graph of the function.

To use the Second Derivative Test we must compute the function’s second derivative. We start with the first derivative that we found in part (a):

\begin{align*} f'(x) &= 6x^2 -6x -12 \\[12px] f”(x) &= \dfrac{d}{dx} \left[6x^2 -6x -12 \right] \\[8px] &=12x – 6 \end{align*}

For the Second Derivative Test, we must determine whether $f”(c)$ is positive or negative (or zero) for each of our critical points. Recall from part (a) that our critical points are $c = -1, \, 2$:

**(I)** Let’s look first at $c = -1$, and determine whether $f”(-1)$ is positive or negative:

\begin{align*} f”(-1) &= 12(-1) – 6 \\[8px] &= -18 < 0 \end{align*} That is, the curve is concave down at $x = -1$, and so this critical point is a local maximum.

**(II)** Next let’s look first at $c = 2$, and determine whether $f”(2)$ is positive or negative:

\begin{align*} f”(2) &= 12(2) – 6 \\[8px] &= 18 > 0 \end{align*} That is, the curve is concave up at $x = 2$, and so this critical point is a local minimum.

The Second Derivative Test is thus a different way or arriving at the conclusions we reached in part (b):

$\bullet$ Since $f”$ is negative $x = -1$, this point is a local maximum. $\quad \cmark$

$\bullet$ Since $f”$ is positive at $x = 2$, this point is a local minimum. $\quad \cmark$

\begin{align*} f'(x) &= 6x^2 -6x -12 \\[12px] f”(x) &= \dfrac{d}{dx} \left[6x^2 -6x -12 \right] \\[8px] &=12x – 6 \end{align*}

For the Second Derivative Test, we must determine whether $f”(c)$ is positive or negative (or zero) for each of our critical points. Recall from part (a) that our critical points are $c = -1, \, 2$:

\begin{align*} f”(-1) &= 12(-1) – 6 \\[8px] &= -18 < 0 \end{align*} That is, the curve is concave down at $x = -1$, and so this critical point is a local maximum.

\begin{align*} f”(2) &= 12(2) – 6 \\[8px] &= 18 > 0 \end{align*} That is, the curve is concave up at $x = 2$, and so this critical point is a local minimum.

The Second Derivative Test is thus a different way or arriving at the conclusions we reached in part (b):

$\bullet$ Since $f”$ is negative $x = -1$, this point is a local maximum. $\quad \cmark$

$\bullet$ Since $f”$ is positive at $x = 2$, this point is a local minimum. $\quad \cmark$

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Question 2: . . . now on a closed interval

Consider again the function from Question 1, $f(x) = 2x^3 -3x^2 -12x +1$.

Find the $x$-values of absolute maximum and the absolute minimum on the interval [0, 3]. **Answer**: Absolute maximum at $x = 0$; absolute minimum at $x = 2$.

From Question 1, we already know the critical points are $c = -1,\, 2$. The first, $c = -1$, is not in our interval and so we won’t consider it further.

To determine the absolute maximum and absolute minimum, we simply have to compute the values of $f$ at the critical point that is in our interval, $c = 2$, as well as at the endpoints, $x = 0$ and 3. The absolute maximum is where $f$ obtains its largest value; the absolute minimum is where $f$ obtains its smallest value.

$\bullet$ $f(2) = 2(2)^3 -3(2)^2 -12(2) + 1 = 16 -12 -24 +1 = -19$

$\bullet$ $f(0) = 1$

$\bullet$ $f(3) = 2(3)^3 -3(3)^2 -12(3) + 1 = 54 -27 -36 +1 = -8$

We thus conclude that, on the interval [0, 3];

$\blacktriangleright$ the absolute maximum occurs at $x = 0$ (where $f(0) = 1$); $\quad \cmark$

$\blacktriangleright$ the absolute minimum occurs at $x = 2$ (where $f(2) = -19$). $\quad \cmark$

Find the $x$-values of absolute maximum and the absolute minimum on the interval [0, 3].

Show/Hide Solution

From Question 1, we already know the critical points are $c = -1,\, 2$. The first, $c = -1$, is not in our interval and so we won’t consider it further.

To determine the absolute maximum and absolute minimum, we simply have to compute the values of $f$ at the critical point that is in our interval, $c = 2$, as well as at the endpoints, $x = 0$ and 3. The absolute maximum is where $f$ obtains its largest value; the absolute minimum is where $f$ obtains its smallest value.

$\bullet$ $f(2) = 2(2)^3 -3(2)^2 -12(2) + 1 = 16 -12 -24 +1 = -19$

$\bullet$ $f(0) = 1$

$\bullet$ $f(3) = 2(3)^3 -3(3)^2 -12(3) + 1 = 54 -27 -36 +1 = -8$

We thus conclude that, on the interval [0, 3];

$\blacktriangleright$ the absolute maximum occurs at $x = 0$ (where $f(0) = 1$); $\quad \cmark$

$\blacktriangleright$ the absolute minimum occurs at $x = 2$ (where $f(2) = -19$). $\quad \cmark$

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Question 3: *f(x)* = *e*^(-*x*) sin(*x*)

Consider the function $f(x) = e^{-x} \sin x$ on the interval $[0, 2\pi]$.**(a)** Find the critical points of $f$.**(b)** Find the points of global maximum and minimum.**(a)** $\dfrac{\pi}{4}, \, \dfrac{5\pi}{4} $**(b)** Global maximum at $\left(\dfrac{\pi}{4}, 0.322 \right)$; global minimum at $\left(\dfrac{5\pi}{4}, -0.0139\right)$

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Solution SummarySolution (a) DetailSolution (b) Detail

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To find the critical points of $f$, we must examine its derivative:
\begin{align*}
f'(x) &= \frac{d}{dx} \left[e^{-x} \sin x \right] \\ \\
&= \left(\frac{d}{dx} e^{-x} \right) \sin x + e^{-x} \left(\frac{d}{dx} \sin x \right) \\ \\
&= \left(-e^{-x}\right) \sin x + e^{-x} (\cos x) \\ \\
&= \left(-e^{-x}\right) (\sin x – \cos x)
\end{align*}

The critical points occur when $f'(c) = 0$. (We could also have critical points where $f'(c)$ is undefined, but $f'(x)$ is defined everywhere.)

$f'(c) = 0$ when $\sin c – \cos c = 0$: \begin{align*} \sin c – \cos c &= 0 \\ \\ \sin c &= \cos c \\ \\ \tan c &= 1 &&\text{[Recall that our interval is }[0, 2\pi]]\\ \\ c &= \frac{\pi}{4}, \, \frac{5\pi}{4} \quad \cmark \end{align*}

The critical points occur when $f'(c) = 0$. (We could also have critical points where $f'(c)$ is undefined, but $f'(x)$ is defined everywhere.)

$f'(c) = 0$ when $\sin c – \cos c = 0$: \begin{align*} \sin c – \cos c &= 0 \\ \\ \sin c &= \cos c \\ \\ \tan c &= 1 &&\text{[Recall that our interval is }[0, 2\pi]]\\ \\ c &= \frac{\pi}{4}, \, \frac{5\pi}{4} \quad \cmark \end{align*}

To find the global (absolute) maximum and minimum, we must compute the values of $f$ at the critical points ($\dfrac{\pi}{4}$ and $\dfrac{5\pi}{4}$) and at the endpoints (0 and $2\pi$):

$\bullet$ $f\left(\dfrac{\pi}{4}\right) = e^{-\frac{\pi}{4}} \sin \left(\dfrac{\pi}{4}\right) = e^{-\frac{\pi}{4}} \dfrac{1}{\sqrt{2}} \approx 0.322$

$\bullet$ $f\left(\dfrac{5\pi}{4}\right) = e^{-\frac{5\pi}{4}} \sin \left(\dfrac{5\pi}{4}\right) = e^{-\frac{5\pi}{4}} \left(-\dfrac{1}{\sqrt{2}}\right) \approx -0.0139$

$\bullet$ $f(0) = e^0 \sin (0) = 0$

$\bullet$ $f(2\pi) = e^{-2\pi} \sin (2\pi) = 0$

The global maximum therefore occurs at the point $\left(\dfrac{\pi}{4}, 0.322 \right)$, while the global minimum occurs at $\left(\dfrac{5\pi}{4}, -0.0139\right)$. $\quad \cmark$

For comparison purposes, the graph of the function is plotted below.

$\bullet$ $f\left(\dfrac{\pi}{4}\right) = e^{-\frac{\pi}{4}} \sin \left(\dfrac{\pi}{4}\right) = e^{-\frac{\pi}{4}} \dfrac{1}{\sqrt{2}} \approx 0.322$

$\bullet$ $f\left(\dfrac{5\pi}{4}\right) = e^{-\frac{5\pi}{4}} \sin \left(\dfrac{5\pi}{4}\right) = e^{-\frac{5\pi}{4}} \left(-\dfrac{1}{\sqrt{2}}\right) \approx -0.0139$

$\bullet$ $f(0) = e^0 \sin (0) = 0$

$\bullet$ $f(2\pi) = e^{-2\pi} \sin (2\pi) = 0$

The global maximum therefore occurs at the point $\left(\dfrac{\pi}{4}, 0.322 \right)$, while the global minimum occurs at $\left(\dfrac{5\pi}{4}, -0.0139\right)$. $\quad \cmark$

For comparison purposes, the graph of the function is plotted below.

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Question 4: *f(x)* = *x* - ln(*x*)

Consider the function $f(x) = x - \ln x$ on the interval [0.1, 2]. Find the values of $x$ for which $f$ has its global maximum and minimum. **Answer:** Global maximum at $x = 0.1$; global minimum at $x= 1$.

Following the Problem Solving Strategy outlined above:

**1. Compute the function’s derivative,** $f'(x)$: \begin{align*}
f'(x) &= \frac{d}{dx}\left[ x – \ln x \right] \\ \\
&= 1 – \frac{1}{x}
\end{align*}

**2. Find all critical numbers $c$ such that either (i) $f'(c) = 0$ or (ii) $f'(c)$ is undefined. **

$f'(x)$ is defined everywhere on the interval [0.1, 2], so we only have to look for the critical values where $f'(c) = 0$:

\begin{align*} f'(c) = 1 – \frac{1}{c} &= 0 \\ \\ c &= 1 \end{align*}

**4, 5. Compute the values of $f(a)$ and $f(b)$ to see how they compare to the values of $f(c)$.**

We now compare the value of $f$ at this critical point, $c = 1$, with the value of $f$ at each of the endpoints, 0.1 and 2:

$\bullet$ $f(1) = 1 – \ln 1 =1 – 0 = 1$

$\bullet$ $f(0.1) = 0.1 – \ln(0.1) \approx 2.4$

$\bullet$ $f(2) = 2 – \ln 2 \approx 1.31$

Hence the function on the interval [0.1, 2]:

$\blacktriangleright$ has its global maximum at $x = 0.1 \quad \cmark$

$\blacktriangleright$ and its global minimum at $x= 1$. $\quad \cmark$

For comparison, we’ve graphed the function below.

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Show/Hide Solution

Following the Problem Solving Strategy outlined above:

$f'(x)$ is defined everywhere on the interval [0.1, 2], so we only have to look for the critical values where $f'(c) = 0$:

\begin{align*} f'(c) = 1 – \frac{1}{c} &= 0 \\ \\ c &= 1 \end{align*}

We now compare the value of $f$ at this critical point, $c = 1$, with the value of $f$ at each of the endpoints, 0.1 and 2:

$\bullet$ $f(1) = 1 – \ln 1 =1 – 0 = 1$

$\bullet$ $f(0.1) = 0.1 – \ln(0.1) \approx 2.4$

$\bullet$ $f(2) = 2 – \ln 2 \approx 1.31$

Hence the function on the interval [0.1, 2]:

$\blacktriangleright$ has its global maximum at $x = 0.1 \quad \cmark$

$\blacktriangleright$ and its global minimum at $x= 1$. $\quad \cmark$

For comparison, we’ve graphed the function below.

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[hide solution]

Question 5: Find *k* such that ... (Based on an actual exam question)

Find the value of $k$ such that $f(x) = x + \dfrac{k}{x}$ has a local minimum at $x = 3$. **Answer: ** $k=9$

We want to find the value of $k$ such that $f$ has a critical point at $x = 3$. That is, we need to choose $k$ such that $f'(3) = 0$. Let’s start by computing $f'(x)$:

\begin{align*} f'(x) &= \frac{d}{dx} \left[ x + \frac{k}{x}\right] \\ \\ &= 1 – \frac{k}{x^2} \end{align*} We want $f'(3) = 0$:

\begin{align*} f'(3) &= 1 – \frac{k}{(3)^2} = 0 \\ \\ 1 – \frac{k}{9} &= 0\\ \\ k &= 9 \quad \cmark \end{align*}

To verify that $f$ has a local minimum (rather than a maximum) at $x = 3$, we can use the Second Derivative Test. We start with the first derivative, now setting $k = 9$: \begin{align*} f'(x) &= 1 – \frac{9}{x^2} \\ \\ f”(x) &= -(-2)\frac{9}{x^3} \\ \\ &= \frac{18}{x^3} \end{align*}

So $f”(3) > 0$, and hence by the Second Derivative Test $x = 3$ is a local minimum (as opposed to a local maximum) when $k=9$. For comparison, the graph of the function with $k=9$ is below.

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Show/Hide Solution

We want to find the value of $k$ such that $f$ has a critical point at $x = 3$. That is, we need to choose $k$ such that $f'(3) = 0$. Let’s start by computing $f'(x)$:

\begin{align*} f'(x) &= \frac{d}{dx} \left[ x + \frac{k}{x}\right] \\ \\ &= 1 – \frac{k}{x^2} \end{align*} We want $f'(3) = 0$:

\begin{align*} f'(3) &= 1 – \frac{k}{(3)^2} = 0 \\ \\ 1 – \frac{k}{9} &= 0\\ \\ k &= 9 \quad \cmark \end{align*}

To verify that $f$ has a local minimum (rather than a maximum) at $x = 3$, we can use the Second Derivative Test. We start with the first derivative, now setting $k = 9$: \begin{align*} f'(x) &= 1 – \frac{9}{x^2} \\ \\ f”(x) &= -(-2)\frac{9}{x^3} \\ \\ &= \frac{18}{x^3} \end{align*}

So $f”(3) > 0$, and hence by the Second Derivative Test $x = 3$ is a local minimum (as opposed to a local maximum) when $k=9$. For comparison, the graph of the function with $k=9$ is below.

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[hide solution]

Question 6: Find *a* & *b* such that. . . (Based on an actual exam question)

Find values $a$ and $b$ so that the function $f(x) = ax e^{-bx}$ has a local maximum at the point (2, 4). **Answer:** $a = 2e, \, b = \dfrac{1}{2}$

We need to find two unknowns, $a$ and $b$. Fortunately, the question is giving us two facts: (I) $f$ has a local maximum at $x = 2$, and (II) $f(2) = 4$. Let’s start with fact (I):

(I) For $f$ to have a local maximum at $x = 2$, we must have $f'(2) = 0$. So let’s start by taking the derivative: \begin{align*} f'(x) &= \frac{d}{dx} \left[ ax e^{-bx}\right] \\ \\ &= \left(\frac{d}{dx} ax \right) e^{-bx} + ax \left(\frac{d}{dx} e^{-bx} \right) \\ \\ &= a e^{-bx} + ax(-b) e^{-bx} \\ \\ &= ae^{-bx} (1 -bx) \end{align*} We want $f'(2) = 0$:

$$f'(2) = ae^{-2b} (1 -2b) = 0$$ Now, we cannot have $a = 0$ or the entire function $f(x) = 0$. So instead we must have

\begin{align*} (1-2b) &= 0 \\ \\ b &= \frac{1}{2} \quad \cmark \end{align*}

(II) Now that we know $b = \dfrac{1}{2}$, let’s use our second fact to determine $a$. We know $f(2) = 4$:

\begin{align*} f(x) &= ax e^{-\frac{x}{2}} \\ \\ f(2) &= 2a e^{-1} = 4 \\ \\ 2a &= 4e \\ \\ a &= 2e \quad \cmark \end{align*} So $a = 2e$ and $b = \frac{1}{2}$, such that $f(x) = (2e)e^{-\frac{x}{2}} = 2 e^{\left(1-\frac{x}{2}\right)}$.

For comparison, we’ve graphed the function with these values of $a$ and $b$.

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Show/Hide Solution

We need to find two unknowns, $a$ and $b$. Fortunately, the question is giving us two facts: (I) $f$ has a local maximum at $x = 2$, and (II) $f(2) = 4$. Let’s start with fact (I):

(I) For $f$ to have a local maximum at $x = 2$, we must have $f'(2) = 0$. So let’s start by taking the derivative: \begin{align*} f'(x) &= \frac{d}{dx} \left[ ax e^{-bx}\right] \\ \\ &= \left(\frac{d}{dx} ax \right) e^{-bx} + ax \left(\frac{d}{dx} e^{-bx} \right) \\ \\ &= a e^{-bx} + ax(-b) e^{-bx} \\ \\ &= ae^{-bx} (1 -bx) \end{align*} We want $f'(2) = 0$:

$$f'(2) = ae^{-2b} (1 -2b) = 0$$ Now, we cannot have $a = 0$ or the entire function $f(x) = 0$. So instead we must have

\begin{align*} (1-2b) &= 0 \\ \\ b &= \frac{1}{2} \quad \cmark \end{align*}

(II) Now that we know $b = \dfrac{1}{2}$, let’s use our second fact to determine $a$. We know $f(2) = 4$:

\begin{align*} f(x) &= ax e^{-\frac{x}{2}} \\ \\ f(2) &= 2a e^{-1} = 4 \\ \\ 2a &= 4e \\ \\ a &= 2e \quad \cmark \end{align*} So $a = 2e$ and $b = \frac{1}{2}$, such that $f(x) = (2e)e^{-\frac{x}{2}} = 2 e^{\left(1-\frac{x}{2}\right)}$.

For comparison, we’ve graphed the function with these values of $a$ and $b$.

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Question 7: A particle's motion

A particle starts at time $t = 0$ and moves on a number line so that its position at time $t$ is given by

$$x(t) = (t -2)^3 (t-6) $$**(a)** When is the particle at rest?**(b)** When is the particle moving to the right?**(c)** When does the particle change direction?**(d)** What is the farthest to the left of the origin that the particle moves?**(a)** $t = 2, \, 5$**(b)** $t > 5$**(c)** $t = 5$**(d)** -27

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$$x(t) = (t -2)^3 (t-6) $$

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Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) DetailSolution (d) Detail

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This question is asking the same thing as, “when is the particle’s position instantaneously not changing?”, which in turn is the same question as, “when is the derivative of $x(t)$ zero?”.

\begin{align*} x'(t) &= \frac{d}{dt} \left[ (t -2)^3 (t-6)\right] \\ \\ &= \left(\frac{d}{dt}(t -2)^3 \right) (t – 6) + (t-2)^3 \left( \frac{d}{dt}(t-6) \right) \\ \\ &= 3(t-2)^2 (t-6) + (t-2)^3(1) \\ \\ &= (t-2)^2\left[3(t-6) + (t-2) \right] \\ \\ &= (t-2)^2\left[ 3t -18 + t – 2 \right] \\ \\ &= (t-2)^2 \left[4t – 20 \right] \\ \\ &= 4 (t-2)^2 (t – 5) \end{align*} We want to find when $x'(t) = 0$:

\begin{align*} x'(t) &= 0 \\ \\ 4 (t-2)^2 (t – 5) &= 0 \\ \\ t &= 2, \, 5 \quad \cmark \end{align*}

\begin{align*} x'(t) &= \frac{d}{dt} \left[ (t -2)^3 (t-6)\right] \\ \\ &= \left(\frac{d}{dt}(t -2)^3 \right) (t – 6) + (t-2)^3 \left( \frac{d}{dt}(t-6) \right) \\ \\ &= 3(t-2)^2 (t-6) + (t-2)^3(1) \\ \\ &= (t-2)^2\left[3(t-6) + (t-2) \right] \\ \\ &= (t-2)^2\left[ 3t -18 + t – 2 \right] \\ \\ &= (t-2)^2 \left[4t – 20 \right] \\ \\ &= 4 (t-2)^2 (t – 5) \end{align*} We want to find when $x'(t) = 0$:

\begin{align*} x'(t) &= 0 \\ \\ 4 (t-2)^2 (t – 5) &= 0 \\ \\ t &= 2, \, 5 \quad \cmark \end{align*}

This question is asking the same thing as, “when is the particle’s position increasing?”, which in turn is the same question as, “when is the derivative of $x(t)$ positive?”.

We determined in part (a) that $$x'(t) = 4 (t-2)^2 (t – 5)$$ We want to know when $x'(t) > 0$. Since $(t-2)^2$ is always positive, $x'(t)$ is positive when $(t – 5)$ is positive, or $$t > 5 \quad \cmark$$

We determined in part (a) that $$x'(t) = 4 (t-2)^2 (t – 5)$$ We want to know when $x'(t) > 0$. Since $(t-2)^2$ is always positive, $x'(t)$ is positive when $(t – 5)$ is positive, or $$t > 5 \quad \cmark$$

We know that the particle *could* change directions at the times we found in part (a), when it is momentarily at rest: $t = 2, \, 5$. But to see if it actually does, we need to examine the direction it is moving during the three intervals (i) $t < 2$; (ii) $2 < t < 5$; and (iii) $t < 5$.

As usual, we compute the value of $x'(t)$ at some convenient values of $t$ for each of the three intervals:

$$\bullet\, x'(1) = 4(-1)^2(-4) = -16 \text{, so $x'(t)$ is negative for $t < 2$.}$$

$$\bullet\, x'(3) = 4(1)^2(-2) = -8 \text{, so $x'(t)$ is negative for $2 < t < 5$.}$$

$$\bullet\, x'(6) = 4(4)^2(1) = 64 \text{, so $x'(t)$ is positive for $5 < t$.}$$

So the particle changes direction only once, from moving to the left ($x'(t)$ < 0: its position decreases) to moving to the right ($x'(t) > 0$: its position increases) at $t = 5$. $\quad \cmark$

As usual, we compute the value of $x'(t)$ at some convenient values of $t$ for each of the three intervals:

$$\bullet\, x'(1) = 4(-1)^2(-4) = -16 \text{, so $x'(t)$ is negative for $t < 2$.}$$

$$\bullet\, x'(3) = 4(1)^2(-2) = -8 \text{, so $x'(t)$ is negative for $2 < t < 5$.}$$

$$\bullet\, x'(6) = 4(4)^2(1) = 64 \text{, so $x'(t)$ is positive for $5 < t$.}$$

So the particle changes direction only once, from moving to the left ($x'(t)$ < 0: its position decreases) to moving to the right ($x'(t) > 0$: its position increases) at $t = 5$. $\quad \cmark$

This question is asking the same thing as, “what is the minimum value of $x(t)$?”. We identified the critical points of $x(t)$ in part (a). In part (c), we found that at $t = 5$ the particle switches from moving to the left to moving to the right, and so that instant marks its most negative position.

Hence its position furthest to the left is $$x(5) = (3)^3(-1) = -27 \quad \cmark$$

Hence its position furthest to the left is $$x(5) = (3)^3(-1) = -27 \quad \cmark$$

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