# Difference between revisions of "Math 22 Concavity and the Second-Derivative Test"

## Formal Definition of Concavity

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


## Test for Concavity

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


## Guidelines for Applying the Concavity Test

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


Exercises: Find the second derivative of ${\displaystyle f}$ and discuss the concavity of its graph.

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

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

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

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

## 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 ${\displaystyle (c,f(c))}$ is a point of inflection of the graph of ${\displaystyle f}$,
then either ${\displaystyle f''(c)=0}$ or ${\displaystyle f''(c)}$ is undefined.


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

Therefore, ${\displaystyle (0,0)}$ is a point of inflection

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

Therefore, ${\displaystyle (0,10)}$ and ${\displaystyle (1,9)}$ are points of inflection.

## The Second-Derivative Test

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


Exercises: Find all relative extrema of

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

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