# Find the intervals of concavity and the inflection points of f(x) = e(\tan^{-1}) - x and f(x) =...

## Question:

Find the intervals of concavity and the inflection points of {eq}f(x) = e(\tan^{-1}) - x {/eq} and {eq}f(x) = \frac{x^2 - 4}{x^2 + 4}. {/eq}

## Concavity and Inflection points

A function can be concave up or concave down, we can know the concavity that has a function with the second derivative test, the sign of the second derivative will determine the concavity. Similarly, the point where the concavity changes are known as the inflection points.

a. We have the function

{eq}f(x)= e(\tan^{-1} (x) ) - x \\ {/eq}

Differentiating the function

{eq}f'(x)={\frac {{\rm e}}{{x}^{2}+1}}-1 \\ f''(x)=-2\,{\frac {{\rm e}x}{ \left( {x}^{2}+1 \right) ^{2}}} \\ {/eq}

{eq}f''(x)=0 {/eq} when {eq}x=0 {/eq}

{eq}\begin{array}{r|D{.}{,}{5}} Interval & {-\infty<x<0} & {0<x<\infty} \\ \hline Test \space{} value & \ x=-1 & \ x=1 \\ Sign \space{} of \ f'' (x) & \ f'' (-1)>0 & \ f'' (1)<0 \\ Conclusion & concave \space up & concave \space down \\ \end{array} \\ {/eq}

Inflection point:

{eq}(0,0 ) \\ {/eq}

b. We have the function

{eq}f(x) = \frac{x^2 - 4}{x^2 + 4} \\ {/eq}

Differentiating the function

{eq}f'(x)=16\,{\frac {x}{ \left( {x}^{2}+4 \right) ^{2}}} \\ f''(x)=-16\,{\frac {3\,{x}^{2}-4}{ \left( {x}^{2}+4 \right) ^{3}}} \\ {/eq}

{eq}f''(x)=0 {/eq} when {eq}x=-\frac{2\sqrt{3} }{3} \\ x= \frac{2\sqrt{3} }{3} \\ {/eq}

{eq}\begin{array}{r|D{.}{,}{5}} Interval & {-\infty<x<-\frac{2\sqrt{3} }{3} } & {-\frac{2\sqrt{3} }{3} <x<\frac{2\sqrt{3} }{3} } & {\frac{2\sqrt{3} }{3} <x<\infty } \\ \hline Test \space{} value & \ x=-2 & \ x=0 & \ x=2 \\ Sign \space{} of \ f'' (x) & \ f'' (-2)<0 & \ f'' (0)>0 & \ f''(2)<0 \\ Conclusion & concave \space down & concave \space up & concave \space down \\ \end{array} \\ {/eq}

Inflection points:

{eq}(\frac{2\sqrt{3} }{3} , -\frac{1}{2} ) \\ (-\frac{2\sqrt{3} }{3} , -\frac{1}{2} ) \\ {/eq}