Answer to Question #159569 in Calculus for sdfD

Question

given (x^2)/x^2-4

Find all critical values for 𝑔(𝑥). Show all work and explain

b. Find 𝑔 ′′(𝑥). Then, use your answers from part (a) and the 2nd derivative test to determine if each critical value represents a relative maximum, minimum, or neither. Show all work and explain

c. On what interval(s), if any, is 𝑔(𝑥) concave down. Show all work and explain

Answer 1

$g(x) = \cfrac{x^2}{x^2-4}$

a) Find all critical values:

Critical values is values where first derivative is equal to zero ($g\prime(x)= 0$ )

So find firs derivative:

$g\prime(x) = (\cfrac{x^2}{x^2-4})\prime = \cfrac{2x(x^2-4)-2x^3}{(x^2-4)^2}= -\cfrac{8x}{(x^2-4)^2}$

Find critical values:

$g\prime(x) = 0\\ -\cfrac{8x}{(x^2-4)^2} = 0 => x_0= 0$

So we have one critical value.

b) Relative maximum or minimum:

If $g\prime\prime(x_0) > 0,$ then $x_0$ - is relative minimum

if $g\prime\prime(x_0) < 0,$ then $x_0$ - is relative maximum.

So find $g\prime\prime(x)$ :

$g\prime\prime(x) = \cfrac{-8(x^2-4)^2 + 32x^2(x^2-4)}{(x^2-4)^4} = \cfrac{-8(x^2-4)+32x}{(x^2-4)^3}$

$g\prime\prime(x_0) = -\cfrac{1}{2} < 0$, so $x_0$ is relative maximum

c) Where $g(x)$ is concave down:

For all $x \in (-\infin, -2) \cup(2,+\infin)$ $g\prime\prime(x) > 0$, so on this intervals function is concave down.