Recall that for an even function, $f(-x) = f(x)$, while for an odd function $f(-x) = -f(x)$. We thus need to determine how $f(-x)$ compares to $f(x)$:

\begin{align*} f(-x) &= \frac{(-x)^3 + 1}{(-x)^2} \\ \\ &= \frac{-x^3 + 1}{x^2} \end{align*}

If the function were even, $f(-x)$ would equal $f(x) = \dfrac{x^3 + 1}{x^2}$, which it does not.

If the function were odd, $f(-x)$ would equal $-f(x) = \dfrac{-\left(x^3 + 1\right)}{x^2}$, which it does not.

Hence the function is neither even nor odd. $\quad \cmark$

To find where the function is increasing and decreasing, we must investigate its derivative, $f'(x)$. Rather than working with the function as written and using the quotient rule, we find it easier to rewrite it as $f(x) = x + \dfrac{1}{x^2}$.

\begin{align*} f(x) &=x + \dfrac{1}{x^2}\\ \\ f'(x) &= 1 – \frac{2}{x^3} \end{align*}

We see immediately that there is one critical point where $f'(c)$ is undefined, at $c = 0$.
To find any others, we set $f'(c) = 0$: \begin{align*} f'(c) = 1 – \frac{2}{c^3} &= 0 \\ \\ 1 &= \frac{2}{c^3} \\ \\ c^3 &= 2 \\ \\ c &= \sqrt[3]{2} \end{align*}

So the critical points are $c = 0, \sqrt[3]{2}$.

We now investigate whether the function is increasing or decreasing in the three regions (i) $x < 0$; (ii) $0 < x < \sqrt[3]{2}$; and (iii) $\sqrt[3]{2} < x$.

$\bullet$ (i) For $x < 0$, $f'(x) > 0$. (Try, for example, $f'(-1) = 1 – \dfrac{2}{(-1)^3} = 3$.)
$\bullet$ (ii) For $0 < x < \sqrt[3]{2}$, $f'(x) < 0$. (Try, for example, $f'(1) = 1 - \dfrac{2}{(1)^3} = -1$.)
$\bullet$ (iii) For $x > \sqrt[3]{2}$, $f'(x) > 0$. (Try, for example, $f'(2) = 1 – \dfrac{2}{(2)^3} = 1 – \dfrac{1}{4} = \dfrac{3}{4}$.)

We summarize our results using the same graphic we used in Max & Min problems: Hence $f(x)$ is increasing on $(-\infty, 0)$ and $(\sqrt[3]{2}, \infty)$, and decreasing on $(0, \sqrt[3]{2})$. $\quad \cmark$

To find where the function is concave up and concave down, we must investigate its second derivative, $f”(x)$:

\begin{align*} f'(x) &= 1 – \frac{2}{x^3} \\ \\ f”(x) &= 0 – \frac{(2)(-3)}{x^4} \\ \\ &= \frac{6}{x^4} \end{align*}

Hence $f”(x)$ is positive for all $x$ (except at $x = 0$, where the function is undefined): the function is concave up for $(-\infty, 0)$ and $(0, \infty)$. It is never concave down. $\quad \cmark$

To sketch the curve, let’s note first that the function is undefined at $x = 0$. Investigating what happens as we approach $x = 0$ from the left and from the right:
$$\bullet \lim_{x \to 0^-}\left[x + \frac{1}{x^2} \right] = 0 + \lim_{x \to 0^-}\frac{1}{x^2} = \infty$$
$$\bullet \lim_{x \to 0^+}\left[x + \frac{1}{x^2} \right] = 0 + \lim_{x \to 0^+}\frac{1}{x^2} = \infty$$
Hence we can sketch the first part of the curve as shown, remembering that we found in parts (b) and (c) that the curve is concave up everywhere (except at $x = 0$), and is increasing we approach the $y$-axis from the left, and decreasing as we leave the $y$-axis going to the right. Next, let’s determine if there are any $x$-intercepts, which would occur when $y = \dfrac{x^3+1}{x^2}= 0$. The function is undefined for $x = 0$, so we only need to determine when (if) the numerator equals 0:

\begin{align*} x^3+1 &= 0\\ \\ x^3 &= -1 \\ x &= -1 \end{align*} Hence there is an $x$-intercept at $x = -1$. We add this knowledge to our sketch as shown. Let’s next consider what happens to $f(x) = x + \dfrac{1}{x^2}$ as $x$ becomes very large in the negative direction $(x \to -\infty)$, and becomes very large in the positive direction $(x \to \infty)$:
$\bullet$ As $x \to -\infty$, $\dfrac{1}{x^2} \to 0$, and so $y = f(x)$ becomes very close to $y = x$.
$\bullet$ As $x \to \infty$, $\dfrac{1}{x^2} \to 0$, and so again $y = f(x)$ becomes very close to $y = x$, shown as a dotted line in the figure below. Finally, we found in part (b) that $f(x)$ has a local minimum at $x = \sqrt[3]{2}$: it is decreasing on the interval $(0, \sqrt[3]{2})$ and increasing on $(\sqrt[3]{2}, \infty)$ Putting the pieces together, we sketch this graph: For comparison, the actual graph is shown below; the second one is more zoomed-out.

To find the slope of the tangent line to the curve, we examine the function’s derivative:

\begin{align*} f'(x) &= \frac{d}{dx} \left[x^{\frac{4}{3}} + 4x^{\frac{1}{3}} \right] \\ \\ &= \frac{4}{3} x^{\frac{1}{3}} + 4\left(\frac{1}{3} \right)x^{-\frac{2}{3}}\\ \\ &= \frac{4}{3} \frac{(x+1)}{x^\frac{2}{3}} \end{align*}

The tangent line is horizontal when the curve’s slope is zero ($f'(x) = 0$). We can see immediately that this occurs when $x = -1$.

To fully specify the point where the curve has zero slope, we need to find the $y$-value when $x = -1$: \begin{align*} y &= (-1)^{\frac{4}{3}} + 4(-1)^{\frac{1}{3}} \\ \\ &= 1 + 4(-1) = -3 \end{align*} Hence the curve has a horizontal tangent at the point $(-1, -3)$. $\quad \cmark$

In part (a) we found that the function’s derivative is $$f'(x) = \frac{4}{3} \frac{(x+1)}{x^\frac{2}{3}} $$ The tangent line is vertical when the derivative does not exist (when it is infinite). We see immediately that this occurs when $x = 0$.

When $x=0$, $y = f(0) = 0$. So the curve has a vertical tangent line at the point (0,0). $\quad \cmark$

Recall from part (a) that the function’s derivative is $$f'(x) = \frac{4}{3} \frac{(x+1)}{x^\frac{2}{3}} $$ The critical points are thus $c = -1$ and $c = 0$.

We thus need to examine the three regions (i) $x < -1$; (ii) $-1 < x < 0$; and (iii) $0 < x$.

$\bullet$ (i) For $x < -1$, $f'(x) <0$, since the numerator is negative while the denominator is positive.

$\bullet$ (ii) For $-1< x < 0$, $f'(x) > 0$, since the numerator is positive and the denominator is positive.

$\bullet$ (iii) For $ x > 0$, $f'(x) > 0$, since again the numerator is positive and the denominator is positive. Hence there is a local minimum at $x = -1$.

To find the absolute maximum and minimum, we must check the endpoints to see how they compare to the local minimum at $x = -1$:

$\bullet$ $f(-1) = (-1)^{\frac{4}{3}} + 4(-1)^\frac{1}{3} = 1 + 4(-1) = -3$
$\bullet$ $f(-8) = (-8)^\frac{4}{3} + 4(-8)\frac{1}{3} = 16 + 4(-2) = 8$
$\bullet$ $f(8) = 8^\frac{4}{3} + 4(8)^\frac{1}{3} = 16 + 4(2) = 24$


Hence the absolute minimum is the point (-1 ,-3) and the absolute maximum is the point (8, 24). $\quad \cmark$

To determine concavity we examine the function’s second derivative. We begin with our result from part (a) for the first derivative. Let’s start with the equation we obtained right after taking the first derivative, before we did any simplifying, since taking the second derivative is then easier: \begin{align*} f'(x) &= \frac{4}{3} x^{\frac{1}{3}} + \frac{4}{3} x^{-\frac{2}{3}} \\ \\ f”(x) &= \left(\frac{4}{3}\right)\left(\frac{1}{3}\right)x^{-\frac{2}{3}} + \left(\frac{4}{3}\right) \left(-\frac{2}{3} \right)x^{-\frac{5}{3}} \\ \\ &= \frac{4}{9} \left[ \frac{x -2}{x^{\frac{5}{3}}} \right] \end{align*}

The numerator changes sign at $x = 2$, while the denominator changes sign at $x = 0$. We thus see that
$\bullet$ For $x < 0$, $f''(x) > 0$ since both the numerator and denominator are negative.
$\bullet$ For $0 < x < 2$, $f''(x) < 0$ since the numerator is negative while the denominator is positive.
$\bullet$ For $x>2$, $f”(x) > 0$ since both the numerator and denominator are positive.
Hence the function is concave down on the interval (0, 2). $\quad \cmark$

Let’s summarize what we’ve found in the earlier parts about the function on the interval [-8, 8]:
$\bullet$ The function’s tangent is vertical at $x=0$, so the graph itself is vertical there.
$\bullet$ The graph has an absolute minimum at (-1, -3).
$\bullet$ The graph has an absolute maximum at (8, 24).
$\bullet$ The graph is concave down on (0,2), and concave up on the other intervals.

We’ve captured these features in the figure shown. Filling in the rest of the graph can result in something like the figure below. A computer-generated graph of the function is shown below for comparison.

To find where the maxima and minima lie, we must examine the derivative of the function:

\begin{align*} f'(x) &= \frac{d}{dx} \left[ x + \sin x \right] \\ \\ &= 1 + \cos x \end{align*} To find the critical points, we set $f'(c) = 0$:

\begin{align*} f'(c) &= 0 \\ 1+ \cos c &= 0 \\ \cos c &= -1 &&\text{[Recall that }-\dfrac{\pi}{2} \le x \le \dfrac{3\pi}{2}]\\ c &= \pi \end{align*}

Our only critical point is at $c = \pi$. But notice that since the most negative $\cos x$ can become is -1, $f'(x) = 1 + \cos x$ is never negative. Hence $x = \pi$ is neither a maximum nor a minimum. (Thinking ahead to part (c), note however that $f(x)$ has a horizontal tangent at $x = \pi$.)

Let’s consider the endpoints of the interval:
$\bullet$ $f(-\dfrac{\pi}{2}) = -\dfrac{\pi}{2} + \sin \left(-\dfrac{\pi}{2}\right) = -\dfrac{\pi}{2} – 1$

$\bullet$ $f(\dfrac{3\pi}{2}) = \dfrac{3\pi}{2} + \sin \left(\dfrac{3\pi}{2}\right) = \dfrac{3\pi}{2} -1 $


Hence the maximum point is $\left(\dfrac{3\pi}{2}, \dfrac{3\pi}{2} -1\right)$, and the minimum point is $\left(-\dfrac{\pi}{2}, -\dfrac{\pi}{2} – 1\right)$ $\quad \cmark$

To find the point(s) of inflection, we must examine the second derivative. We begin with the first derivative that we found in part (a): $f'(x) = 1 + \cos x$.

\begin{align*} f'(x) &= 1 + \cos x \\ \\ f”(x) &= -\sin x \\ \\ \end{align*} The points of inflection occur when $f”(x) = 0$:
$$ -\sin x = 0$$

On our interval $\left[-\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$, $-\sin x = 0$ at

$$x = 0, \, \pi$$

We thus need to examine $f”(x)$ in the three regions: (i) $-\dfrac{\pi}{2} < x < 0$; (ii) $0 < x < \pi$; (iii) $\pi < x < \dfrac{3\pi}{2}$:
$\bullet$ (i) For $-\dfrac{\pi}{2} < x < 0$, $-\sin x > 0$.
$\bullet$ (ii) For $0 < x < \pi$, $-\sin x < 0$.
$\bullet$ (iii) For $\pi < x < \dfrac{3\pi}{2}$, $-\sin x > 0$. Hence we see that both $x = 0$ and $x = \pi$ are points of inflection.

We need the $y$-value of those points in order to answer the question as asked:
$\bullet$ $f(0) = 0 + \sin 0 = 0$
$\bullet$ $f(\pi) = \pi + \sin \pi = \pi$

The coordinates of the points of inflection are therefore (0, 0) and $(\pi, \pi)$. $\quad \cmark$

Let’s summarize what we know from the preceding parts of this question:
$\bullet$ The function has a minimum at $\left(-\frac{\pi}{2}, -\frac{\pi}{2} – 1\right)$.
$\bullet$ The function has maximum at $\left(\frac{3\pi}{2}, \frac{3\pi}{2} -1\right)$.
$\bullet$ The function has a horizontal tangent at $(\pi, \pi)$.
$\bullet$ The function is concave down on the interval $(0, \pi)$, and concave up on the other two intervals.

We’ve sketched these facts in this figure. Joining the pieces together leads to something like this figure. The figure below is a computer-generated graph of the function for comparison.

To determine when $f(x) = 0$, let’s factor the expression: \begin{align*} f(x) &= \cos^2 x + 2\cos x = 0\\ &= \cos x(\cos x + 2) = 0 \end{align*}
The term in parentheses is always positive, so has no zeros. (Remember that the most negative $\cos x$ can be is $-1$.)

Hence the function is zero only when $\cos x = 0$. On the interval $[0, 2\pi]$, this occurs for $$x = \dfrac{\pi}{2} \text{ and } \dfrac{3\pi}{2}\quad \cmark$$

Recall $f(x) = \cos^2 x + 2\cos x$. Then

\begin{align*} f'(x) &= 2\cos x (-\sin x) + 2(-\sin x) \\ \\ &= (-2\sin x )(\cos x + 1) \\ \end{align*}

The critical points, such that $f'(c) = 0$, occur when either term in parentheses is zero on our interval $[0, 2\pi]$: \begin{equation*} \begin{aligned}[c] -2\sin c &= 0 \\ c &= 0, \pi, 2\pi \end{aligned} \qquad \text{or} \qquad \begin{aligned}[c] \cos c + 1 &= 0 \\ \cos c &= -1 \\ c &= \pi \end{aligned} \end{equation*} Our critical points are thus $c = 0$, $\pi$, and $2\pi$.

To determine where $f'(x)$ is positive and where it is negative, we note that $(-2\sin x)$ is negative on the interval $(0, \pi)$, and positive on the interval $(\pi, 2\pi)$. The term $(\cos x + 1)$ is positive on the interval $(0, \pi)$, and positive on the interval $(\pi, 2\pi)$: Hence the product $(-2\sin x)(\cos x + 1)$is negative on the first interval, and positive on the second:

We thus see by the First Derivative Test that the function has a minimum at $x = \pi$. $\quad \cmark$

From part (b), $f'(x) = (-2\sin x )(\cos x + 1)$. Then

\begin{align*} f”(x) &= (-2 \cos x)(\cos x + 1) + (-2\sin x )(-\sin x) \\ \\ &= -2 \cos^2 x – 2 \cos x + 2 \sin^2 x \\ \\ &= 2 \left[ -\cos^2 x – \cos x + \sin^2 x \right] \\ \\ &= 2 \left[ -\cos^2 x – \cos x + (1 – \cos^2 x) \right] \\ \\ &= 2 \left[ -2\cos^2 x – \cos x + 1 \right] \end{align*} Then $f”(x) = 0$ when:

\begin{align*} f”(x) = 0 &= -2\cos^2 x – \cos x + 1 \\ \\ &= (-2 \cos x +1)(\cos x + 1) \end{align*}

Thus $f”(x) = 0$ when either term in parentheses is zero: \begin{equation*} \begin{aligned}[c] (2 \cos x – 1) &= 0 \\ \cos x &= \dfrac{1}{2} \\ x &= \dfrac{\pi}{3}, \dfrac{5\pi}{3} \end{aligned} \qquad \text{or} \qquad \begin{aligned} (\cos x + 1) = 0 \\ \cos x = -1 \\ x = \pi \end{aligned} \end{equation*}

Our critical points are thus $x = \dfrac{\pi}{3}$, $\pi$, and $\dfrac{5\pi}{3}$.

Recall from above that $f”(x) = 2(-2 \cos x +1)(\cos x + 1)$.

To determine where $f'(x)$ is positive and where it is negative, we note that the term $(-2\cos x + 1)$ is negative on the interval $(0, \dfrac{\pi}{3})$ since $\cos x > \dfrac{1}{2}$ there. Similarly, $(-2\cos x + 1)$ is positive on the interval $(\dfrac{\pi}{3}, \dfrac{5\pi}{3})$ since $\cos x < \dfrac{1}{2}$ there. And then on the final interval, $(\dfrac{5\pi}{3}, 2\pi)$, the term $(-2\cos x + 1)$ is again negative since again $\cos x > \dfrac{1}{2}$ there.

By contrast, the term $(\cos x + 1)$ is positive everywhere, except at $x = \pi$ where it is zero. Hence the product is negative on the intervals $(0, \dfrac{\pi}{3})$ and $(\dfrac{5\pi}{3}, 2\pi)$, and positive on the intervals $(\dfrac{\pi}{3}, \pi)$ and $(\pi, \dfrac{5\pi}{3})$: Therefore $f$ is concave up on the intervals $(\dfrac{\pi}{3}, \pi)$ and $(\pi, \dfrac{5\pi}{3})$. $\quad \cmark$

To find where the absolute maximum and absolute minimum occur we must examine $f'(x)$: \begin{align*} f(x) &= \frac{1}{x} + \ln x \\ \\ f'(x) &= -\frac{1}{x^2} + \frac{1}{x} \\ \\ &= \frac{x – 1}{x^2} \end{align*} $f'(c) = 0$ when $c=1$, so that is a critical number. (Note that $c=0$ is not a critical number, since the function is only defined on the interval $[\dfrac{1}{e}, e]$.)

To find the absolute maximum and absolute minimum, we must compute $f(x$) at the critical number, and at the endpoints of the interval, and compare the values: \begin{align*} f(1) &= \frac{1}{1} + \ln(1) = 1 + 0 = 1 \\ \\ f(\frac{1}{e}) &= \frac{1}{\frac{1}{e}} + \ln \left(\frac{1}{e} \right) = e – 1 \\ \\ f(e) &= \frac{1}{e} + \ln(e) = \frac{1}{e} + 1 \\ \\ \end{align*}

Recall $e \approx 2.7$. Hence $1 < \frac{1}{e} + 1 < e - 1$.

Therefore (i) the absolute maximum occurs at $x = \dfrac{1}{e}$, and (ii) the absolute minimum occurs at $x = 1$. $\quad \cmark$

To determine concavity, we must examine $f”(x)$. We begin with the expression for $f'(x)$ that we found in part (a):

\begin{align*} f'(x) &= -\frac{1}{x^2} + \frac{1}{x} \\ \\ f”(x) &= \frac{2}{x^3} – \frac{1}{x^2} \\ \\ &= \frac{2-x}{x^3} \end{align*}

Thus $f”(x) > 0$ for $x < 2$, and $f$ is concave up on the interval $[ \dfrac{1}{e}, 2)$. $\quad \cmark$

See the top sketch. We’ve marked the three points we found in part (b), making sure that the absolute maximum and absolute minimum are correctly shown.
Although it’s tough to see given how little the function grows between $x = 1$ and $x =e$, we have tried to indicate a change from positive to negative concavity at $x = 2$.

Below is the function’s graph as generated by a graphing program, just for comparison.

Another way to state this question is, “Show that the graph of $f$ only crosses the $x$-axis once and only once.” There are two parts to showing this:
(I) The graph has at least one root, and so crosses the $x$-axis at least once. We’ll use the Intermediate Value Theorem to show this, that there exists some value $c$ $(-\infty < c < \infty)$ for which $f(c) = 0$.
(II) We’ll then show (we hope) that the function either never decreases, or never increases, so that it never “turns around and heads back toward zero.” If that’s the case, then it can cross the axis at most once.

We’ll thus have shown that (I) the function crosses the $x$-axis at least once, and (II) it crosses the $x$-axis at most once. Putting those together, we’ll have shown that the function crosses the $x$-axis exactly once, meaning it has exactly one root.

(I) Let’s first show that the function has at least one root.

First note that $\displaystyle{\lim_{x \to -\infty} f(x) = -\infty}$, while $\displaystyle{\lim_{x \to \infty} f(x) = \infty}$. Since $f$ is a continuous function, by the Intermediate Value Theorem there must be some value $c$ $(-\infty < c < \infty)$ for which $f(c) = 0$. That is, $f(x)$ has at least one root, and crosses the $x$-axis at least once.

(II) Let’s next see how the function increases and/or decreases by examining its derivative: \begin{align*} f(x) &= x^3 -6x^2 + 12x – 4 \\ f'(x) &= 3x^2 -12x + 12 \\ &= 3(x^2 – 4x + 4) \\ &= 3(x – 2)^2 \end{align*}

Hence $f'(x) = 0$ at $x = 2$; everywhere else ($x \ne 2$), $f'(x) > 0$. That is, the function increases everywhere except at $x = 2$; it never decreases, never turns-around and heads downward. It can thus cross the $x$-axis at most once. Since $f$ crosses the $x$-axis once, but no more than once, it must have exactly one real root. $\quad \cmark$

We’ll take the same approach we did in part (a), and (I) show that the function has at least one root, and then (II) see how it increases or decreases.

(I) To show that $g(x)$ has at least one root, let’s again use the Intermediate Value Theorem (IVT). To keep things simple, let’s try some values for $x$ for which $\cos x = 0$: $x = -\pi/2$ and $\pi/2$.

\begin{align*} g(-\pi/2) &= 6(-\dfrac{\pi}{2}) – \cos(\dfrac{\pi}{2}) = -3\pi \text{, which is } < 0 \\ g(\pi/2) &= 6(\dfrac{\pi}{2}) - \cos(\dfrac{\pi}{2}) = 3\pi \text{, which is } > 0 \end{align*}

Since $g(x)$ is a continuous function, we then know from the IVT that there exists $c$, $-\dfrac{\pi}{2} < c < \dfrac{\pi}{2}$, such that $g(c) = 0$. Thus $g(x)$ has at least one real root.

(II) To see how the function increases and/or decreases, we examine its derivative: \begin{align*} g(x) &= 6x – \cos x \\ g'(x) &= 6 + \sin x \end{align*}

Since $-1 \le \sin x \le 1$, we know $g'(x) > 0$ for all $x$. That is, $g(x)$ always increases; it never decreases. It can thus cross the $x$-axis at most once: it has at most one real root.

Since $g(x)$ has at least one root, but at most one root, it must have exactly one real root. $\quad \cmark$