Skill: Determine concavity using the second derivative

Concavity tells you how the slope of a graph is changing:

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

To determine concavity:

1. Find $f''(x)$.
2. Find where $f''(x)=0$ or where $f''(x)$ is undefined.
3. Use these values to divide the domain into intervals.
4. Test the sign of $f''(x)$ on each interval.
5. An inflection point occurs only where the concavity changes.

Example: Determine the intervals of concavity and inflection points for

$$f(x)=x^4-4x^3.$$

Step 1: Find the second derivative

First,

$$f'(x)=4x^3-12x^2.$$

Differentiate again:

$$f''(x)=12x^2-24x=12x(x-2).$$

Step 2: Find possible boundary points

Set the second derivative equal to zero:

$$12x(x-2)=0.$$

Therefore,

$$x=0 \quad \text{or} \quad x=2.$$

Since $f$ is a polynomial, its domain is all real numbers, and $f''(x)$ is never undefined. The values $0$ and $2$ divide the domain into three intervals:

$$(-\infty,0), \qquad (0,2), \qquad (2,\infty).$$

Step 3: Test the sign of $f''(x)$

Use $f''(x)=12x(x-2)$:

• On $(-\infty,0)$, choose $x=-1$:

$$f''(-1)=12(-1)(-3)>0.$$

The graph is concave up.

• On $(0,2)$, choose $x=1$:

$$f''(1)=12(1)(-1)<0.$$

The graph is concave down.

• On $(2,\infty)$, choose $x=3$:

$$f''(3)=12(3)(1)>0.$$

The graph is concave up.

Thus,

$$\boxed{\text{Concave up on }(-\infty,0)\text{ and }(2,\infty)}$$

and

$$\boxed{\text{Concave down on }(0,2)}.$$

Step 4: Identify inflection points

At $x=0$, the concavity changes from up to down.

At $x=2$, the concavity changes from down to up.

Therefore, both values give inflection points. Find their coordinates:

$$f(0)=0,$$

so one inflection point is $(0,0)$.

Also,

$$f(2)=2^4-4(2^3)=16-32=-16,$$

so the other inflection point is $(2,-16)$.

Remember: a value where $f''(x)=0$ is only a possible inflection point. It is an actual inflection point only if the concavity changes there.