Math 22 Concavity and the Second-Derivative Test

Formal Definition of Concavity

 Let  $f$  be differentiable on an open interval  $I$ . The graph of  $f$  is
 1. Concave upward on  $I$  when  $f'(x)$  is increasing on the interval.
 2. Concave downward on  $I$  when  $f'(x)$  is decreasing on the interval.

Test for Concavity

 Let  $f$  be a function whose second derivative exists on an open interval  $I$ 
 1. If  $f''(x)>0$  for all  $x$  in  $I$ , then the graph of  $f$  is concave upward on  $I$ .
 2. If  $f''(x)<0$  for all  $x$  in  $I$ , then the graph of  $f$  is concave downward on  $I$ .

Guidelines for Applying the Concavity Test

 1. Locate the  $x$ -values at which  $f''(x)=0$  or  $f''(x)$  is undefined.
 2. Use these  $x$ -values to determine the test intervals.
 3. Determine the sign of  $f'(x)$  at an arbitrary number in each test intervals
 4. Apply the concavity test

Exercises: Find the second derivative of $f$ and discuss the concavity of its graph.

1) $f(x)=x^{3}+2x^{2}$

ExpandSolution:
Step 1: $f'(x)=3x^{2}+4x$ , so $f''(x)=6x=0$
Step 2: So $x=0$ , so the test intervals are $(-\infty ,0)$ and $(0,\infty )$
Step 3: Choose $x=-1$ for the interval $(-\infty ,0)$ , and $x=1$ for the interval $(0,\infty )$ .
Then we have: $f''(-1)=-6<0$ and $f''(1)=6>0$
Step 4: By the concavity test, $f(x)$ is concave up in $(0,\infty )$ and $f(x)$ is concave down in $(-\infty ,0)$

2) $f(x)=x^{4}-2x^{3}+10$

ExpandSolution:
Step 1: $f'(x)=4x^{3}-6x^{2}$ , so $f''(x)=12x^{2}-12x=12x(x-1)=0$
Step 2: So, $x=0$ and $x=1$ , so the test intervals are $(-\infty ,0),(0,1)$ and $(1,\infty )$
Step 3: Choose $x=-1$ for the interval $(-\infty ,0)$ , $x={\frac {1}{2}}$ for the interval $(0,1)$ and $x=2$ for the interval $(1,\infty )$ .
Then we have: $f''(-1)=24>0$ , $f''({\frac {1}{2}})=-3<0$ and $f''(2)=24>0$
Step 4: By the concavity test, $f(x)$ is concave up in $(-\infty ,0)\cup (1,\infty )$ and $f(x)$ is concave down in $(0,1)$

Points of Inflection

 If the graph of a continuous function has a tangent line at a point 
 where its concavity changes from upward to downward (or downward to upward), 
 then the point is a point of inflection.
 
 If  $(c,f(c))$  is a point of inflection of the graph of  $f$ , 
 then either  $f''(c)=0$  or  $f''(c)$  is undefined.

In exercises 1, at $x=0$ , the concavity changes from concave down to concave up, so $(0,f(0))$ is a point of inflection.

Therefore, $(0,0)$ is a point of inflection

In exercises 2, at $x=0$ and $x=1$ , the concavity changes from concave up to concave down and from concave down to concave up, respectively. So, $(0,f(0))$ and $(1,f(1))$ are points of inflection.

Therefore, $(0,10)$ and $(1,9)$ are points of inflection.

The Second-Derivative Test

  Let  $f'(c)=0$ , and let  $f''(x)$  exist on an open interval containing  $c$ ,
 1. If  $f''(c)>0$ , then  $f(c)$  is relative minimum.
 2. If  $f''(c)<0$ , then  $f(c)$  is relative maximum.
 3. If  $f''(c)=0$ , then the test fails. Use the first derivative test.

Exercises: Find all relative extrema of

1) $f(x)=2x^{3}+3x^{2}-5$

ExpandSolution:
Notice, $f'(x)=6x^{2}-6x=6x(x-1)=0$ , then critical numbers are $x=0$ and $x=1$
$f''(x)=12x-6$ , then $f''(0)=-6<0$ and $f''(1)=6>0$
By the second derivative test, $f(0)=-5$ is relative maximum and $f(1)=0$ is relative minimum
Therefore, relative maximum: $(0,-5)$ and relative minimum : $(1,0)$