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This function is symmetric across the $y$-axis, so it is even.

The function is not even, because it is not mirrored across the $y$-axis.

Assuming it continues as shown, the function is periodic, because the unit shown below is repeated:

Additionally, $f(x)$ is odd. In a function with odd symmetry, if we mirror the right-hand portion of the curve (the portion to the right of the $y$-axis) across both the $y$-axis and the $x$-axis, it lines up with the left-hand portion of the curve.

Since reflecting the right-hand portion of the graph across the $y$-axis, then the $x$-axis, gives us $f(x)$, we conclude $f(x)$ is odd.

Since the function is even, we simply reflect the portion shown across the $y$-axis to complete the sketch.

Since the function is odd, to complete the sketch, we reflect the portion shown across the $y$-axis (shown dashed), then the $x$-axis (shown in red).

A function is even if $f(-x)=f(x)$.

$$\begin{aligned} f(-x)&=\frac{(-x)^4-(-x)^6}{e^{(-x)^2}}\\ &=\frac{x^4-x^6}{e^{x^2}}\\ &=f(x) \end{aligned}$$

So, $f(x)$ is even.

For any real number $x$, we will show that $f(x)=f(x+4\pi)$.

$$\begin{aligned} f(x+4\pi)&=\sin(x+4\pi)+\cos\left(\frac{x+4\pi}{2}\right)\\ &=\sin(x+4\pi)+\cos\left(\frac{x}{2}+2\pi\right)\\ &=\sin(x)+\cos\left(\frac{x}{2}\right)\\ &=f(x) \end{aligned}$$

So, $f(x)$ is periodic.

Let's consider $g(x)=\tan(3x)$ and $h(x)=\sin(4x)$ separately. Recall that $\pi$ is the period of tangent.

$$\begin{aligned} \tan X &= \tan(X+\pi)&&\text{for every $X$ in the domain of $\tan X$} \\ \text{Replacing $X$ with $3x$:} \\ \tan(3x)&=\tan(3x+\pi)&&\text{for every $x$ in the domain of $\tan 3x$}\\ \tan(3x)&=\tan\left(3\left(x+\frac{\pi}{3}\right)\right)&&\text{for every $x$ in the domain of $\tan 3x$}\\ g(x)&=g\left(x+\frac{\pi}{3}\right)&&\text{for every $x$ in the domain of $\tan 3x$} \\ \text{So, the period of $g(x)=\tan(3x)$ is $\dfrac{\pi}{3}$.} \\ \\ \text{Similarly, $2\pi$ is the period of sine.} \\ \sin(X)&=\sin(X+2\pi)&&\text{for every $X$ in the domain of $\sin(X)$} \\ \text{Replacing $X$ with $4x$:} \\ \sin(4x)&=\sin(4x+2\pi)&&\text{for every $x$ in the domain of $\sin(4x)$}\\ \sin(4x)&=\sin\left(4\left(x+\frac{\pi}{2}\right)\right)&&\text{for every $x$ in the domain of $\sin(4x)$}\\ h(x)&=h\left(x+\frac{\pi}{2}\right)&&\text{for every $x$ in the domain of $\sin(4x)$} \end{aligned}$$

So, the period of $h(x)=\sin(4x)$ is $\dfrac{\pi}{2}$.

All together, $f(x)=g(x)+h(x)$ will repeat when both $g(x)$ and $h(x)$ repeat. The least common integer multiple of $\dfrac{\pi}{3}$ and $\dfrac{\pi}{2}$ is $\pi$. Since $g(x)$ repeats every $\dfrac{\pi}{3}$ units, and $h(x)$ repeats every $\dfrac{\pi}{2}$ units, they will not both repeat until we move $\pi$ units. So, the period of $f(x)$ is $\pi$.

(a) First, we differentiate. $$f(x) = x^5-x\qquad f'(x)=5x^4-1\qquad f''(x)=20x^3\qquad$$ The function and its first derivative tells us the following:

• $\displaystyle \lim_{x \to \infty} f(x)=\infty$, $\displaystyle \lim_{x \to -\infty} f(x)=-\infty$
• $f'(x)> 0$ (i.e. $f$ is increasing) for $|x|> \dfrac{1}{\sqrt[4]{5}}$
• $f'(x)=0$ (i.e. $f$ has critical points) for $x= \pm\dfrac{1}{\sqrt[4]{5}} \approx \pm 0.67$
• $f'(x)< 0$ (i.e. $f$ is decreasing) for $|x| < \dfrac{1}{\sqrt[4]{5}}$
• $f\left(\pm\dfrac{1}{\sqrt[4]{5}}\right)=\mp\dfrac{4}{5\sqrt[4]{5}}\approx\mp0.53$

This gives us a first idea of the shape of the graph.

We refine this skeleton using information from the second derivative.

• $f''(x)> 0$ (i.e. $f$ is concave up) for $x> 0$,
• $f''(x)=0$ (i.e. $f$ has an inflection point) for $x=0$, and
• $f''(x)< 0$ (i.e. $f$ is concave down) for $x<0$

Thus

• $f$ has no asymptotes
• $f$ has a local maximum at $x=-\dfrac{1}{\sqrt[4]{5}}$ and a local minimum at $x=\dfrac{1}{\sqrt[4]{5}}$
• $f$ has an inflection point at $x=0$
• $f$ is concave down for $x<0$ and concave up for $x>0$

(b) The function $x^5-x+k$ has a root at $x=x_0$ if and only if $x^5-x=-k$ at $x=x_0$. So the number of distinct real roots of $x^5-x+k$ is the number of times the curve $y=x^5-x$ crosses the horizontal line $y=-k$. The local maximum of $x^5-x$ (when $x=-\dfrac{1}{\sqrt[4]{5}}$) is $\dfrac{4}{5\sqrt[4]{5}}$, and the local minimum of $x^5-x$ (when $x=\dfrac{1}{\sqrt[4]{5}}$) is $-\dfrac{4}{5\sqrt[4]{5}}$. So, looking at the graph of $x^5-x$ above, we see that the number of distinct real roots of $x^5-x+k$ is

• 1 when $|k|>\dfrac{4}{5\sqrt[4]{5}}$
• 2 when $|k|=\dfrac{4}{5\sqrt[4]{5}}$
• 3 when $|k|<\dfrac{4}{5\sqrt[4]{5}}$

(a) Writing $p(x) = x^3 - ax^2 = x^2(x-a)$, we see that $p(x) = 0$ at $x=0$ and $x=a$. Further, since $x^2 \ge 0$ for all $x$, we note that $p(x) \le 0$ whenever $x < a$ and $p(x) > 0$ for all $x > a$. A typical such function is plotted below. (b) To find the critical numbers of $p$, we first compute $p'(x)$. Treating $a$ as constant, we have $p'(x) = 3x^2 - 2ax = x(3x - 2a)$. Thus, setting $p'(x) = 0$, we find that $x(3x-2a) = 0$ and thus $x = 0$ and $x = \frac{2a}{3}$ are the critical numbers of $p$. (c) Observe that $p''(x) = 6x - 2a$ and thus $p''(x) = 0$ implies that $x = \frac{a}{3}$. Since $p''(x)$ is linear with positive slope, we see that $p''(x)$ changes sign from negative to positive at $x = \frac{a}{3}$, and thus $p$ is concave down for $x < \frac{a}{3}$ and concave up for $x > \frac{a}{3}$ and $x = \frac{a}{3}$ is the only inflection point of $p$. (d) There is always a critical number of $p$ at $x = 0$. But the other critical number at $x = \frac{2a}{3}$ and the inflection point at $x = \frac{a}{3}$ both obviously depend directly on $a$: as we increase the value of $a$, both the location of the critical number and the inflection point move to the right along with $a$.

(a) The value of the numerator at $x=c$ is $e^{-c} \ne 0$, while the value of the denominator is $0$ at $x=c$. Thus, $\lim_{x \to c^+} \frac{e^{-x}}{x-c} = \infty$ and $\lim_{x \to c^-} \frac{e^{-x}}{x-c} = -\infty$. This makes the line $x=c$ a vertical asymptote for $q$. (b) We can write $q(x)$ as $q(x) = \frac{1}{(x-c)e^x}$ and thus see the denominator increases without bound as $x \to \infty$. Therefore, $\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0$. This makes the line $y=0$ a horizontal asymptote for $q$. As $x \to -\infty$, $e^{-x}$ increases without bound, while $(x-c)$ decreases without bound. Thus, $q$ has an indeterminate form as $x \to -infty$. Applying L'Hopital's Rule gives us $$\lim_{x \to -\infty} \frac{e^{-x}}{x-c} = \lim_{x \to -\infty} \frac{-e^{-x}}{1} = -\infty\,.$$ (c) Using $q(x) = \frac{1}{(x-c)e^x}$, the quotient and product rules along with algebraic simplification show that $$q'(x) = -\frac{(x-c)e^x + e^x}{(x-c)^2\left(e^x\right)^2} = -\frac{x-c+1}{(x-c)^2e^x}\,.$$ The function $q$ has critical points in its domain where $q'(x) = 0$ or $q'(x)$ doesn't exist. Note that $q'(x)$ is undefined at $x=c$, but $c$ is not in the domain of mq. Thus, the only critical points of $q$ occur when $q'(x)=0$. The only way the fraction $q'(x)$ can be $0$ is if the numerator is $0$. Since $x-c+1=0$ when $x=c-1$, the only critical number for $q$ is $x=c-1$. (d) The values of the exponential function $e^x$ and the quadratic $(x-c)^2$ are positive for $x \neq c$, so the sign of $q'(x) = -\frac{x-c+1}{(x-c)^2e^x}$ depends entirely on the sign of $x-(c-1)$. When $x < c-1$ we see that $q'(x) > 0$ and when $x > c-1$ we have $q'(x) < 0$. Thus, $q$ is increasing on $(-\infty, c-1)$ and is decreasing on $(c-1,c)$, so it follows that $q(c-1)$ is a relative maximum value. This value is a local maximum, since the value of $q$ can be made as large as we like as shown in part (b). (e) We've shown that a typical member of the family of functions given by $q(x) = \frac{1}{(x-c)e^x}$ has a vertical asymptote at $x = c$ and a critical number at which a local maximum occurs one unit to the left of $c$ at $x = c-1$. A typical such graph is shown below. Note how $q$ is increasing and decreasing according to our earlier work in (d).