# Determine the open intervals on which the graph is concave up words or concave downward. \\ f(x)...

## Question:

Determine the open intervals on which the graph is concave up words or concave downward.

{eq}f(x) = \frac{x^2 + 6}{x^2 - 4} {/eq}

## Concavity of the Graph

The inflection point is a point where the concavity of the graph changes. The Inflection point is calculated by equating the second derivative with 0.

if {eq}f''(x)<0 {/eq}; concave down

If {eq}f''(x)>0 {/eq}; concave up

Given:

{eq}f(x)= \dfrac{(x^2+6)}{(x^2-4)}\\ {/eq}

Differentiating with respect to x:

{eq}f'(x)= \dfrac{(x^2+6)\dfrac{\mathrm{d} (x^2-4)}{\mathrm{d} x}- (x^2-4)\dfrac{\mathrm{d} (x^2+6)}{\mathrm{d} x}}{(x^2-4)^{2}}\\ f'(x)= \dfrac{2x^3-8x-2x^3-12x}{(x^2-4)^{2}}\\ f'(x)= -\dfrac{20x}{(x^2-4)^{2}}\\ {/eq}

Again differentiating with respect to x and equating with 0 to get the inflection point:

{eq}f''(x)= -(\dfrac{(x^2-4)^{2}\times 20- 20(2(x^2-4))2x}{(x^2-4)^{4}})\\ f''(x)= \dfrac{-20(x^2-4)^{2}+80x(x^2-4)}{(x^2-4)^{4}}\\ f''(x)= (x^2-4)\dfrac{-20(x^2-4)+80x}{(x^2-4)^{4}}\\ f''(x)= \dfrac{-20(x^2-4)+80x}{(x^2-4)^{3}}\\ \dfrac{-20(x^2-4)+80x}{(x^2-4)^{3}}=0\\ -20x^2++80+80x=0,\ \ \ and\ \ \ (x^2-4)^{3}\neq 0\\ -x^2+4+4x=0\\ x^2-4x-4=0\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ a=1,\ b=4,\ c=4\\ {/eq}

Plugging the value:

{eq}x= \dfrac{4\pm\sqrt{16+16}}{2(1)}\\ x= \dfrac{4+\sqrt{32}}{2(1)}\ or\ \dfrac{4-\sqrt{32}}{2(1)}\\ x= 4.83\ or\ -0.83\\ {/eq}

The denominator must not be zero, so

{eq}(x^2-4)^{3} \neq 0 x^2-4\neq 0\\ x^2\neq 4\\ x\neq \pm\sqrt{2} {/eq}

We get the intervals: {eq}(-\infty,-2),\ (-2, -0.83),\ (-0.83,2), (2,4.83)\ (4.83,\infty) {/eq}

For {eq}(-\infty,-2) {/eq}; test value: -3

{eq}f''(-3)= \dfrac{-20((-3)^2-4)+80(-3)}{((-3)^2-4)^{3}}\\ f''(-3)= \dfrac{-180+80-240}{(3^2-4)^{3}}\\ f''(-3)= -\dfrac{340}{5^2}\\ f''(-3)= -\dfrac{340}{25}\\ f''(x)<0;\ concave\ down\\ {/eq}

For{eq}(-2, -0.83) {/eq}; test value: -1

{eq}f''(-1)= \dfrac{-20((-1)^2-4)+80(-1)}{((-1)^2-4)^{3}}\\ f''(-1)= \dfrac{-20+8-80}{(1-4)^{3}}\\ f''(-1)= \dfrac{-92}{(-3)^3}\\ f''(-1)= \dfrac{-92}{-27}\\ f''(-1)= \dfrac{92}{27}\\ f''(x)>0;\ concave\ up\\ {/eq}

For {eq}(-0.83,2) {/eq}; test value: 0

{eq}f''(0)= \dfrac{-20((0)^2-4)+80(0)}{((0)^2-4)^{3}}\\ f''(0)= \dfrac{80}{(-4)^3}\\ f''(0)= \dfrac{80}{-64}\\ f''(x)<0;\ concave\ down\\ {/eq}

For {eq}(2,4.83) {/eq}; test value:3

{eq}f''(3)= \dfrac{-20((3)^2-4)+80(3)}{((3)^2-4)^{3}}\\ f''(3)= \dfrac{-180+80+240}{(3^2-4)^{3}}\\ f''(3)= \dfrac{140}{5^2}\\ f''(3)= \dfrac{140}{25}\\ f''(x)>0;\ concave\ up\\ {/eq}

For {eq}(4.83, \infty) {/eq}; test value:5

{eq}f''(5)= \dfrac{-20((5)^2-4)+80(5)}{((5)^2-4)^{3}}\\ f''(5)= \dfrac{-500+80+400}{(5^2-4)^{3}}\\ f''(5)= -\dfrac{-500+80}{21^3}\\ f''(5)= -\dfrac{20}{9261} f''(x)<0;\ concave\ down\\ {/eq}

Concave up: {eq}(-2, -0.83) \cup (2,4.83) {/eq}

Concave down: {eq}(-\infty,-2) \cup (-0.83,2) \cup (4.83, \infty) {/eq}

There also exits vertical asymptote at {eq}x=\pm2 {/eq}