7.6 Sketching examples

You'll find the intervals of increase and decrease. These will give you a basic outline of the behaviour of the function. Use concavity to refine your picture.

(a) $(-\infty,3]$

(b) $f(x)$ in increasing on $(-\infty,2)$ and decreasing on $(2,3)$. There is a local maximum at $x=2$ and a local minimum at the endpoint $x=3$.

(c) $f(x)$ is always concave down and has no inflection points.

(d) $(3,0)$

(e)

The local maximum is also a global maximum.

The square mark is the inflection point, and the closed dot is the local and global maximum.

The sign of the first derivative is determined entirely by the numerator, but the sign of the second derivative depends on both the numerator and the denominator.

The square mark marks the inflection point.

The function is odd.

The function is continuous at $x=0$, but its derivative is not.

(a) One branch of the function, the exponential function $e^x$, is continuous everywhere. So $f(x)$ is continuous for $x<0$. When $x \geq 0$, $f(x)=\dfrac{x^2+3}{3(x+1)}$, which is continuous whenever $x \neq -1$ (so it's continuous for all $x >0$). So, $f(x)$ is continuous for $x>0$. To see that $f(x)$ is continuous at $x=0$, we see:

Hence $f(x)$ is continuous at $x=0$, so $f(x)$ is continuous everywhere.
(b) i.
$f(x)$ is increasing for $x<0$ and $x>1$, decreasing for $0<x<1$, has a local max at $(0,1)$, and has a local min at $\left(1,\frac{2}{3}\right)$.
ii.
$f(x)$ is concave upwards for all $x\ne 0$.
iii.
The $x$–axis is a horizontal asymptote as $x\rightarrow-\infty$.

(c)

Since you aren't asked to find the intervals of concavity exactly, sketch the intervals of increase and decrease, and turn them into a smooth curve. You might not get exactly the intervals of concavity that are given in the solution, but there should be the same number of intervals as the solution, and they should have the same positions relative to the local extrema.

Use intervals of increase and decrease, concavity, and asymptotes to sketch the curve.

(a) Increasing: $(-1,1)$ decreasing: $(-\infty,-1)\cup (1,\infty)$
concave up: $(-\sqrt{3},0) \cup (\sqrt{3},\infty)$ concave down: $(-\infty,-\sqrt{3}) \cup (0 ,\sqrt{3})$
inflection points: $x=\pm\sqrt{3}, 0$

(b) The local and global minimum of $f(x)$ is at $(-1,\frac{-1}{\sqrt{e}})$, and the local and global maximum of $f(x)$ is at $(1,\frac{1}{\sqrt{e}})$.

(c) In the graph below, square marks are inflection points, and solid dots are extrema.

Although the function exhibits a certain kind of repeating behaviour, it is not periodic.

Local maxima occur at $x=\frac{2\pi}{3}+2\pi n$ for all integers $n$, and local minima occur at $x=-\frac{2\pi}{3}+2\pi n$ for all integers $n$. Inflection points occur at every integer multiple of $\pi$.

The period of this function is $2\pi$. So, it's enough to graph the curve $y=f(x)$ over the interval $[-\pi,\pi]$, because that figure will simply repeat.

Use trigonometric identities to write $f''(x)=-4(4\sin^2 x + \sin x -2)$. Then you can find where $f''(x)=0$ by setting $y=\sin x$ and solving $0=4y^2+y-2$.

Below is the graph $y=f(x)$ over the interval $[-\pi,\pi]$. The sketch of the curve over a larger domain is simply a repetition of this figure.

On the interval $[0,\pi]$, the maximum value of $f(x)$ is $6$ and the minimum value is $-2$.

Let $a=\sin^{-1}\left(\dfrac{-1+\sqrt{33}}{8}\right)\approx 0.635\approx0.2\pi$ and $b=\sin^{-1}\left(\dfrac{-1-\sqrt{33}}{8}\right)\approx -1.003\approx -0.3\pi$. The points $-\pi-b$, $b$, $a$, and $\textcolor{blue}{\pi-a}$ are inflection points.

There is one point where the curve is continuous but has a vertical tangent line.

The closed dot is the local minimum, and the square marks are inflection points at $x=-1$ and $x=-2\pm\sqrt{1.5}$. The graph has horizontal asymptotes $y=0$ as $x$ goes to $\pm \infty$.

Use $\displaystyle\lim_{x \rightarrow - \infty}f'(x)$ to determine $\displaystyle\lim_{x \rightarrow - \infty}f(x)$.

(a) decreasing for $x<0$ and $x>2$, increasing for $0 < x < 2$, minimum at $(0,0)$, maximum at $(2,2)$.
(b) concave up for $x<2-\sqrt{2}$ and $x>2+\sqrt{2}$, concave down for $2-\sqrt 2 < x < 2+\sqrt 2$, inflection points at $x = 2\pm \sqrt{2}$.

(c)$\infty$

(d)

Square marks indicate inflection points, and closed dots indicate local extrema.

Once you have the graph of a function, reflect it over the line $y=x$ to graph its inverse. Be careful of the fact that $f(x)$ is only defined in this problem for $x \geq 0$.

(a)

There are no inflection points or extrema, except the endpoint $(0,1)$.

(b)

There are no inflection points or extrema, except the endpoint $(1,0)$.

(c) The domain of $g$ is $(0,1]$. The range of $g$ is $[0,\infty)$.

(d) $g'(\half)=-2$

(a)

Local maximum at $x=-\frac{1}{\sqrt[4]{5}}$; local minimum at $x=\frac{1}{\sqrt[4]{5}}$; inflection point at the origin; concave down for $x<0$ ; concave up for $x>0$.

(b) The number of distinct real roots of $x^5-x+k$ is:

For (a), don't be intimidated by the new names: we can graph these functions using the methods learned in this section.

For (b), remember that to define an inverse of a function, we need to restrict the domain of that function to an interval where it is one-to-one. Then to graph the inverse, we can simply reflect the original function over the line $y=x$.

For (c), set $y(x)=\cosh^{-1}(x)$, so $\cosh(y(x))=x$. The differentiate using the chain rule. To get your final answer in terms of $x$ (instead of $y$), use the identity $\cosh^2(y)-\sinh^2(y)=1$.

(a)

(b) For any real $x$, define $\sinh^{-1}(x)$ to be the unique solution of $\sinh(y)=x$. For every $x\in[1,\infty)$, define $\cosh^{-1}(x)$ to be the unique $y\in[0,\infty)$ that obeys $\cosh(y)=x$.

(c) $\ds\diff{}{x}\{\cosh^{-1}(x)\}=\dfrac{1}{\sqrt{x^2-1}}$