Question

use a table to graph (f(x)=x^{4}+8x^{2}-12) on a separate sheet of paper. then estimate the x - coordinates at which relative maxima and relative minima occur. the relative maximum occurs near (x=) select choice. the relative minimum occurs near (x=) select choice. need help with this question? get a hint

Explanation:

Step1: Find the derivative

Let $y = x^{4}+8x^{2}-12$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $y'=4x^{3}+16x=4x(x^{2}+4)$.

Step2: Set the derivative equal to zero

Set $y' = 0$. Since $x^{2}+4>0$ for all real $x$, then $4x(x^{2}+4)=0$ when $4x = 0$, so $x = 0$.

Step3: Use the second - derivative test

Find the second - derivative $y''=12x^{2}+16$. When $x = 0$, $y''(0)=16>0$. So the function has a relative minimum at $x = 0$. And since this is the only critical point (where $y'=0$), there is no relative maximum.

Answer:

The relative maximum occurs near $x=$ None; The relative minimum occurs near $x = 0$