Graphs & curve sketching

$f^{\prime }(x)$ is a polynomial and therefore defined for all values.

$f^{\prime }(x)=0$ when the graph shown crosses the $x$-axis.

• Occurs at $x=-2,1,$ and $3$, which are therefore critical points of $f(x)$.

• At $x=-2$: $f^{\prime }(x)$ changes from negative to positive $\to$ relative min at $x=-2$.

• At $x=1$: $f^{\prime }(x)$ changes from positive to negative $\to$ relative max at $x=1$.

• At $x=3$: $f^{\prime }(x)$ changes from negative to positive $\to$ relative min at $x=3$.

You may also justify using the 2nd derivative test by reading whether $f^{\prime }(x)$ is increasing or decreasing at each critical point:

• At $x=-2$: $f^{\prime }(x)$ is increasing, so $f^{\prime \prime }(-2)>0\to$ relative min

• At $x=1$: $f^{\prime }(x)$ is decreasing, so $f^{\prime \prime }(1)<0\to$ relative max

• At $x=3$: $f^{\prime }(x)$ is increasing, so $f^{\prime \prime }(3)>0\to$ relative min

Only graphs B and C show extrema in this order, so A and D can be eliminated.

Notice that graph B consists of straight-line segments with sharp corners at $x=-2,1,3$. Since corners are not differentiable, the graph of $f^{\prime }$ for choice B would be undefined. However, the given graph of $f^{\prime }(x)$ is a polynomial that shows each point to equal $0$.

So graph B can be eliminated, and graph C is the only option.

Answer: $x=3$ (choice D)

Let’s analyze each point:

At $x=-3$:

• $f(-3)<0$ (graph is below the $x$-axis).
• $f^{\prime }(-3)=0$ (relative min at $x=-3$).
  • Since $f(-3)<f^{\prime }(-3)$, this answer choice can be eliminated.

At $x=-1$:

• $f(-1)=0$ (graph crosses the $x$-axis)
• $f^{\prime }(-1)>0$ ($f$ is increasing).
  • Since $f(-1)<f^{\prime }(-1)$, this answer choice can also be eliminated.

At $x=1$:

• $f(1)>0$
• $f^{\prime }(1)=0$ (relative max at $x=1$).
• $f^{\prime \prime }(1)<0$ (concave down)
  • Since $f^{\prime \prime }(1)<f^{\prime }(1)<f(1)$, this choice can be eliminated.

At $x=3$:

• $f(3)=0$
• $f^{\prime }(3)<0$ ($f$ is decreasing).
• $f^{\prime }(3)>0$ ($f$ changes concavity at $x=2.75$).
  • Since $f^{\prime }(3)<f(3)<f^{\prime \prime }(3)$, this is the correct point.

• Vertical asymptotes: $x=1$ and $x=-1$
• Horizontal asymptote: $y=0$
• $x$-intercepts (when $y=0$): $(0,0)$
• $y$-intercept (when $x=0$): $(0,0)$
• Domain: $x\ne -1$ and $x\ne 1$

With the quotient rule,

$$\begin{array}{rl}{f}^{\prime }(x)& =\frac{(x^2-1)(1)-(x)(2x)}{(x^2-1{)}^2}\\ & =\frac{x^2-1-2x^2}{(x^2-1{)}^2}\\ & =\frac{-x^2-1}{(x^2-1{)}^2}\end{array}$$

Now identify where $f^{\prime }(x)=0$ or is undefined.

$f^{\prime }(x)$ is undefined when $x=-1$ and $x=1$. These are vertical asymptotes where $f(x)$ is undefined, so they are not critical points of $f$.

$f^{\prime }(x)=0$ when $-x^2-1=0$, which has no real solutions. So $f$ has no relative extrema.

Choose a test point in each interval of the domain (split by the vertical asymptotes at $x=-1$ and $x=1$) and determine the sign of $f^{\prime }(x)$. Instead of a chart, the signs are depicted as a diagram:

Differentiate $f^{\prime }(x)$ to find $f^{\prime \prime }(x)$:

$$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{(x^2-1{)}^2(-2x)-(-x^2-1)(2(x^2-1)(2x))}{(x^2-1{)}^4}\\ & =\frac{-2x(x^2-1)-4x(x^2-1)(-x^2-1)}{(x^2-1{)}^4}\end{array}$$

Simplify by factoring $-2x(x^2-1)$ in the numerator:

$$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{-2x(x^2-1)[1+2(-x^2-1)]}{(x^2-1{)}^4}\\ & =\frac{-2x(x^2-1)(-2x^2-1)}{(x^2-1{)}^4}\end{array}$$

Potential inflection points occur where $f^{\prime \prime }(x)=0$ (and $f$ is defined). Here, $f^{\prime \prime }(x)=0$ when $x=0$ only.

Test the sign of $f^{\prime \prime }(x)$ on each interval. Even though $x=-1$ and $x=1$ cannot be inflection points, they still split the domain and should appear on the sign diagram.

Combine what you know about intercepts, asymptotes, increasing/decreasing behavior, and concavity. For example, on $(1,\infty )$, $f(x)$ is decreasing and concave up, and a small curve has been drawn to depict this.