# The function f(x)= \frac {x^3}{x^2-16} defined on the interval [-19,16]. Enter points, such as...

## Question:

The function {eq}\displaystyle f(x)= \frac {x^3}{x^2-16} {/eq} defined on the interval {eq}[-19,16] {/eq}. Enter points, such as inflection points in ascending order, i.e. smallest {eq}x {/eq} values first. Enter intervals in ascending order also.

a) What are {eq}f(x) {/eq} two vertical asymptotes?

b) What is the inflection point?

c) What is the concave up on the region for {eq}f(x) {/eq}?

## Insert context header here:

A Function with one independent variable has inflection point where its second derivative is zero, remember, if the second derivative is positive the function is concave up and if the second derivative is negative, the function is concave down. Also, vertical asymptote exists where the denominator is zero.

## Answer and Explanation:

The function is:

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

Its first derivative is: {eq}\displaystyle \ f'(x)= 3\,{\frac {{x}^{2}}{{x}^{2}-16}}-2\,{\frac {{x}^{4}}{ \left( {x}^{2}- 16 \right) ^{2}}} \\ \displaystyle \ f'(x)= {\frac {{x}^{2} \left( {x}^{2}-48 \right) }{ \left( {x}^{2}-16 \right) ^{2}}} \\ \displaystyle \ f'(x)= {\frac {{x}^{2} \left( {x}^{2}-48 \right) }{ \left( x-4 \right) ^{2} \left( x+4 \right) ^{2}}} \\ {/eq}

Its second derivative is: {eq}\displaystyle \ f''(x)= 6\,{\frac {x}{{x}^{2}-16}}-14\,{\frac {{x}^{3}}{ \left( {x}^{2}-16 \right) ^{2}}}+8\,{\frac {{x}^{5}}{ \left( {x}^{2}-16 \right) ^{3}}} \\ \displaystyle \ f''(x) = 32\,{\frac {x \left( {x}^{2}+48 \right) }{ \left( {x}^{2}-16 \right) ^ {3}}} \\ \displaystyle \ f''(x) = 32\,{\frac {x \left( {x}^{2}+48 \right) }{ \left( x-4 \right) ^{3} \left( x+4 \right) ^{3}}} \\ {/eq}

Inflection point at: {eq}\displaystyle \ f''(x) = 0 {/eq} or {eq}\ f''(x) = indeterminate {/eq} so,

{eq}\displaystyle 0= \ f''(x) \; \Rightarrow \; 0= 32\,{\frac {x \left( {x}^{2}+48 \right) }{ \left( x-4 \right) ^{3} \left( x+4 \right) ^{3}}} \; \Leftrightarrow \; x= 0 \\ {/eq} and, second derivative is indeterminate at (denominator equal zero): {eq}\displaystyle \left( x-4 \right) ^{3} \left( x+4 \right) ^{3}=0 \; \Leftrightarrow \; x=-4 \; \text{&} \; x=4 \\ {/eq}

The concavity intervals are:

{eq}\begin{array} \; \text{Interval} \; & { -19 \leq x < -4 } & { -4 < x < 0 } & { 0 < x < 4 } & { 4 \; < \; x \; \leq \; 16 } \\ \hline Test \space{} value & { x = -11 } & { x = -0.5 } & { x = 0.5 } & { x = 6 } \\ Test & { f''( -11 ) = - < 0 } & { f''( -0.5 ) = + > 0 } & { f''( 0.5 ) = - < 0 } & { f''( 6 ) = + > 0 } \\ Conclusion & { concave \; downward } & { concave \; upward } & { concave \; downward } & { concave \; upward } \\ \end{array} \\ \therefore \boxed { \text{ concave upward intervals are} \; \ ( -4 , 0 )\cup( 4 , 16 ]} \\ \therefore \boxed { \text{ concave downward interval is} \; \ [ -19 , -4 )\cup( 0 , 4 )} \\ {/eq}

Graph

a) Vertical asymptotes at function's denominator is zero:

{eq}\displaystyle \ f(x) = {\frac {{x}^{3}}{ \left( x-4 \right) \left( x+4 \right) }} \\ \displaystyle \; \Rightarrow \; 0= \left( x-4 \right) \left( x+4 \right) \; \Leftrightarrow \; x=-4 \; \text{&} \; x=4 \\ {/eq}

b) The inflection point is: {eq}(0, f(0)) \; \Rightarrow \; (0,0) {/eq}

c) The concave up on the region for {eq}\displaystyle \ f(x) {/eq} is: {eq}\displaystyle ( -4 , 0 )\cup( 4 , 16 ] {/eq}