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 : $(1,0)$

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