Given the functiom y=x^4 -8x^2+16. on which intervals is the function increasing

Question

asked Jun 27, 2020 177k views

1 Answer

Stoiczek · Answer 1 · 2020-07-02T22:21:20+0000

Answer:

$x\in (-2,0)\cup (2,\infty)$

Explanation:

Find the derivative of the function
y=x^4 -8x^2 +16:

$y'=4x^3 -8\cdot 2x\\ \\y'=4x^3 -16x$

The function is increasing when
y'>0, so solve the inequality

$4x^3-16x>0\\ \\4x(x^2-4)>0\\ \\4x(x-2)(x+2)>0\\ \\x\in (-2,0)\cup (2,\infty)$

You can see from the graph that the function increases for
$x\in (-2,0)\cup (2,\infty)$