# Difference between revisions of "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) $f(x)=x^{3}+2x^{2}$ Solution:
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) $f(x)=x^{4}-2x^{3}+10$ Solution:
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) $f(x)=2x^{3}+3x^{2}-5$ Solution:
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 : $(1,0)$ 