11.6: Curve Sketching

Solution

Start with \(\textcolor{brown}{f'(x)=12x^3-12x^2-24x}\)
\(12x^3-12x^2-24x=0 \longrightarrow 12x(x^2-x-2)=0 \longrightarrow 12x(x-2)(x+1)=0 \longrightarrow x=0,2,-1\)
\(f'\) always exists, so these are all of the critical points.

Next we see \(\textcolor{purple}{f''(x)=36x^2-24x-24} \)
\(36x^2-24x-24=0 \longrightarrow 12(3x^2-2x-2)=0 \longrightarrow x=\displaystyle{ \frac{1 \pm \sqrt{7}}{3} }\) (quadratic formula)
\(f''\) always exists, so these are our two inflection points.

Now we put these all in order on the sign chart again:

Plugging each of these critical points in give us:

\(\textcolor{red}{f(-1)=3\cdot(-1)^4-4\cdot(-1)^3-12\cdot(-1)^2+16=11} \)
\(\textcolor{red}{f(0)=3\cdot 0^4-4\cdot 0^3-12\cdot 0^2+16=16} \)
\(\textcolor{red}{f(2)=3\cdot 2^4-4\cdot 2^3-12\cdot 2^2+16=-16} \)

The inflection points look a bit too ugly to find the exact height of, and there are no asymptotes since this is again a polynomial. Lets sketch:

Solution

First derivative: \(\textcolor{blue}{g'(x)=\displaystyle{ \frac{(x^2-4)\cdot 2- 2x\cdot(2x)}{(x^2-4)^2}=\frac{-2x^2-8}{(x^2-4)^2}}}\)

\( g'(x)=0 \longrightarrow -2x^2-8=0 \longrightarrow \emptyset \)

\( g'(x)=DNE \longrightarrow (x^2-4)^2=0 \longrightarrow x=\pm 2 \)

CP: \(x=2,-2\)

Second derivative: \(\textcolor{brown}{g''(x)=\displaystyle{ \frac{(x^2-4)^2\cdot (-4x)-(-2x^2-8)\cdot \left[ 2\cdot(x^2-4)\cdot2x\right]}{(x^2-4)^4}=\frac{4x^3+48x }{(x^2-4)^3}}}\)

\( g''(x)=0 \longrightarrow 4x^3+48x=0 \longrightarrow 4x(x^2+12)=0 \longrightarrow x=0 \)

\( g''(x)=DNE \longrightarrow (x^2-4)^3=0 \longrightarrow x=\pm 2 \)

IP: \(x=0,2,-2\)

Once again we make a sign chart and determine which shape goes to each region:

Plugging in \(x=0\) is simple: \(\textcolor{red}{g(0)=\displaystyle{\frac{2\cdot0}{0^2-4}=0}} \)

\(g(2)\) and \(g(-2)\) do not exist, but limits on each side determine how our graph acts near each.

\(\displaystyle{ \lim_{x\to 2^+} g(x)=\frac{4}{0^+}=+\infty}\) and \(\displaystyle{ \lim_{x\to 2^-} g(x)=\frac{4}{0^-}=-\infty}\)
so vertical asymptote at \(x=2\)

\(\displaystyle{ \lim_{x\to -2^+} g(x)=\frac{-4}{0^-}=+\infty}\) and \(\displaystyle{ \lim_{x\to -2^-} g(x)=\frac{-4}{0^+}=-\infty}\)
so vertical asymptote at \(x=-2\)

While we are at it, lets check for horizontal asymptotes:

\( \displaystyle{\lim_{x\to \infty}g(x)=\lim_{x\to \infty} \frac{2x}{x^2}=\lim_{x\to \infty} \frac{2}{x}=\frac{2}{\infty}=0} \)
\( \displaystyle{\lim_{x\to -\infty}g(x)=\lim_{x\to -\infty} \frac{2x}{x^2}=\lim_{x\to -\infty} \frac{2}{x}=\frac{2}{-\infty}=0}\)

So horizontal asymptote at \(y=0\)

Finally we draw all this and connect the dots:

Solution

\(h'(x)\displaystyle{ =\frac{-5e\cdot e^x}{(e^x-e)^2} }\)
Setting top and bottom equal to zero, we get critical points: \(x=1\)

\(h''(x)\displaystyle{ =\frac{5e\cdot e^x(e^x+e)}{(e^x-e)^3} }\)
Setting top and bottom equal to zero, we get inflection points: \(x=1\)

Lots of algebra but just quotient rule then set the top and bottom equal to zero

Now a sign chart:

\(h(1)=\displaystyle{\frac{2e^1+3e}{e^1-e}=\frac{5e}{0}}=DNE\) so there are no points to plot

\(\displaystyle{\lim_{x\to 1^+} h(x)=\frac{5e}{0^+}=+\infty}\) and \(\displaystyle{\lim_{x\to 1^-} h(x)=\frac{5e}{0^-}=-\infty} \)
Therefore there is a vertical asymptote at \(x=1\)


\(\displaystyle{\lim_{x\to \infty} h(x)\stackrel{LH}{=}\lim_{x\to \infty} \frac{2e^x}{e^x}=\lim_{x\to \infty} 2=2}\)

\(\displaystyle{\lim_{x\to -\infty} h(x)=\frac{2\cdot 0+3e}{0-e}=\frac{3e}{-e}=-3}\)
Notice: for \(x\to -\infty\), you cannot use L'Hopital's Rule since it is not an indeterminate form

This means there are horizontal asymptotes at \(y=2\) and \(y=-3\)

Solution

\(\displaystyle{y'= \frac{4-4x^2}{\sqrt{4-2x^2}}}\), means we have critical points: \(x=\pm 1, \pm \sqrt{2}\)

\(\displaystyle{y''= \frac{8x^3-24x}{(4-2x^2)^{\frac{3}{2}}} }\), means we have inflection points; \(x=0,\pm \sqrt{2}, \pm \sqrt{3}\)

The original function only exists when \(4-2x^2\ge 0\), or \(x\in \left[ -\sqrt{2},\sqrt{2}\right]\)

So we ignore everything outside of this domain for our sign chart:

We plug in each critical and inflection point:

\(y(\sqrt{2})=\sqrt{2}\cdot \sqrt{ 4- 2\cdot (\sqrt{2})^2}=0\)
\(y(1)=1\cdot \sqrt{ 4- 2\cdot 1^2}=\sqrt{2}\)
\(y(0)=0\cdot \sqrt{ 4- 2\cdot 0^2}=0\)
\(y(-1)=-1\cdot \sqrt{ 4- 2\cdot (-1)^2}=-\sqrt{2}\)
\(y(-\sqrt{2})=-\sqrt{2}\cdot \sqrt{ 4- 2\cdot (-\sqrt{2})^2}=0\)

This is continuous on its domain, so no vertical asymptotes.

Since the domain is closed, there is no limit to either infinity (so no horizontal asymptotes either)

We just plot each point and connect the dots:

Solution

\(\displaystyle{f'(x)=\frac{1}{1+\left(x^{-2}\right)^2}\cdot (-2x^{-3})=-\frac{2x}{x^4+1}}\)
So we have critical point \(x=0\)

\(\displaystyle{f''(x)=\frac{6x^4-2}{(x^4+1)^2} } \)
So there are inflection points \(x=\pm \sqrt[4]{3}\)

Even though \(x=0\) is not in the domain of \(f\), the derivative exists on either side of it so we will include it in the sign chart (and explore exactly what happens with limits shortly).

We have a discontinuity at \(x=0\), but what kind?

\(\displaystyle{ \lim_{x\to 0} f(x)=\tan^{-1}\left(\frac{1}{0^+}\right)=\tan^{-1}(+\infty)=\frac{\pi}{2}}\)

\(\displaystyle{ f\left( \pm \sqrt[4]{\frac{1}{3}}\right)=\tan^{-1} \left( \frac{1}{\left(\pm 3^{-\frac{1}{4}}\right)^2}\right)=\tan^{-1}\left(\sqrt{3}\right)=\frac{\pi}{3}}\)

There is a removable discontinuity at \(x=0\), forming a hole at \(\displaystyle{\left(0,\frac{\pi}{2}\right)}\)

\( \displaystyle{\lim_{x\to \infty} f(x)=\tan^{-1}\left(\frac{1}{\infty}\right)=\tan^{-1}(0)=0}\)
\( \displaystyle{\lim_{x\to -\infty} f(x)=\tan^{-1}\left(\frac{1}{\infty}\right)=\tan^{-1}(0)=0} \)
This means we have a horizontal asymptote at \(y=0\)