Skill: Sketch curves using derivative information

The key connections are:

• If $f'(x)>0$, the graph is increasing.
• If $f'(x)<0$, the graph is decreasing.
• A change in $f'(x)$ from positive to negative gives a local maximum.
• A change in $f'(x)$ from negative to positive gives a local minimum.
• If $f''(x)<0$, the graph is concave down.
• If $f''(x)>0$, the graph is concave up.

Use these facts together with intercepts and end behavior to sketch the curve.

Example

Sketch the curve

$$f(x)=x^3-3x^2+2.$$

1. Find the intercepts

For the $y$-intercept, set $x=0$:

$$f(0)=2.$$

So the $y$-intercept is $(0,2)$.

For the $x$-intercepts, solve $f(x)=0$:

$$x^3-3x^2+2=0.$$

Factoring gives

$$x^3-3x^2+2=(x-1)(x^2-2x-2).$$

Thus,

$$x=1,\qquad x=1-\sqrt{3},\qquad x=1+\sqrt{3}.$$

The graph crosses the $x$-axis at these three points.

2. Use $f'(x)$ to find increasing and decreasing intervals

Differentiate:

$$f'(x)=3x^2-6x=3x(x-2).$$

The critical numbers occur where $f'(x)=0$:

$$3x(x-2)=0 \quad\Longrightarrow\quad x=0,\ 2.$$

Test the sign of $f'(x)$ on the intervals determined by $0$ and $2$:

$$\begin{array}{c|c|c} \text{Interval} & \text{Sign of }f'(x) & \text{Behavior}\\ \hline (-\infty,0) & + & \text{increasing}\\ (0,2) & - & \text{decreasing}\\ (2,\infty) & + & \text{increasing} \end{array}$$

Therefore, the graph increases, then decreases, then increases again.

Find the function values at the critical numbers:

$$f(0)=2,\qquad f(2)=8-12+2=-2.$$

Since $f'(x)$ changes from positive to negative at $x=0$, there is a local maximum at

$$(0,2).$$

Since $f'(x)$ changes from negative to positive at $x=2$, there is a local minimum at

$$(2,-2).$$

3. Use $f''(x)$ to determine concavity

Differentiate again:

$$f''(x)=6x-6.$$

Set $f''(x)=0$:

$$6x-6=0 \quad\Longrightarrow\quad x=1.$$

Check the sign of $f''(x)$:

• If $x<1$, then $f''(x)<0$, so the graph is concave down.
• If $x>1$, then $f''(x)>0$, so the graph is concave up.

The concavity changes at $x=1$, so there is an inflection point. Its $y$-coordinate is

$$f(1)=1-3+2=0.$$

Thus, the point of inflection is

$$(1,0).$$

4. Determine the end behavior

The leading term is $x^3$. Therefore,

$$x\to-\infty \implies f(x)\to-\infty,$$

and

$$x\to\infty \implies f(x)\to\infty.$$

So the graph goes down on the far left and up on the far right.

5. Put the information together

Important points and features are:

• $x$-intercepts: $(1-\sqrt3,0)$, $(1,0)$, and $(1+\sqrt3,0)$
• $y$-intercept and local maximum: $(0,2)$
• Local minimum: $(2,-2)$
• Inflection point: $(1,0)$
• Increasing on $(-\infty,0)$ and $(2,\infty)$
• Decreasing on $(0,2)$
• Concave down for $x<1$
• Concave up for $x>1$

Sketch a smooth cubic curve that starts low on the left, rises to the local maximum $(0,2)$, falls through the inflection point $(1,0)$ to the local minimum $(2,-2)$, and then rises to the right.