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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Analytical

Concavity and Inflection Points

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Concavity and Inflection Points

While the first derivative tells us about where a function is increasing and decreasing, the second derivative tells us about a functions Concavity. Concavity is the "U" shape of the graph (either up or down).

Concavity

Assume f′(x)f'(x) is differentiable on the interval (a,b)(a, b)
  • If f′′(x)>0f^{\prime\prime}(x)>0 then f(x)f(x) is concave up
  • If f′′(x)<0f^{\prime\prime}(x)<0 then f(x)f(x) is concave down


Inflection Points

An inflection point of a function f(x)f(x) is a point ,in its domain, at which there is a change in concavity in the graph of f(x)f(x)

Wize Tip
An infection point can occur where f(x)′′=0f(x)''=0 or where the second derivative is not defined.

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Example: Concavity and Inflection Points

Determine the intervals of concavity and locate any inflection points for f(x)=ln⁡(1+x2)f\left(x\right)=\ln\left(1+x^2\right)

NOTE: There is an error in the video. From 4:35-4:50, the presenter mistakenly swaps which intervals are concave up and concave down. Otherwise, everything in the video and written solution is correct.
f′(x)=2x1+x2f'\left(x\right)=\dfrac{2x}{1+x^2}


f′′(x)=2(1+x2)−2x(2x)(1+x2)2f''\left(x\right)=\dfrac{2\left(1+x^2\right)-2x\left(2x\right)}{\left(1+x^2\right)^2}


=2−2x2(1+x2)2=\dfrac{2-2x^2}{\left(1+x^2\right)^2}


f′′(x)=0   at   x=±1f''\left(x\right)=0\ \ \ at\ \ \ x=\pm1


f′′(−2)=−625f′′(0)=212f′′(2)=−625<0>0<0Concave DownConcave UpConcave Down\begin{array}{c|c|c} f''(-2)=\dfrac{-6}{25}& f''(0)=\dfrac{2}{1^2}& f''(2)=\dfrac{-6}{25}\\ \\ <0& >0& <0\\ \\ \text{Concave Down}& \text{Concave Up}& \text{Concave Down}\\ \end{array}


Concave down for (−∞,−1)∪(1,∞)(-\infin,-1) \cup(1,\infin)

Concave up for (−1,1)(-1,1)

(−1,ln⁡2) and (1,ln⁡2)(-1,\ln2) \ \text{and} \ (1,\ln2) are inflection points
Find the equation of the tangent line to the function f(x)=ln⁡ xx f(x)=\frac{\ln\ x}{x} at its inflection point.