Math 22 Concavity and the Second-Derivative Test

Formal Definition of Concavity

 Let $f$ be differentiable on an open interval Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle I}
. The graph of $f$ is
 1. Concave upward on Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle I}
 when $f'(x)$ is increasing on the interval.
 2. Concave downward on Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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: $f''(-1)=24>0$, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f''({\frac {1}{2}})=-3<0} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (c,f(c))}
 is a point of inflection of the graph of $f$, then either Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f''(c)=0}
 or $f''(c)$ is undefined.

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