Question

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

Answer 1

`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)`