Question

For what values of x is `f(x)=2x^4+4x^3+2x^2-2` concave or convex?

Answer 1

See solution below

Explanation:

Given function:

`f(x)=2x^4+4x^3+2x^2-2`

`f'(x)=8x^3+12x^2+4x`

`f''(x)=24x^2+24x+4`

The function will be concave when `f''(x)\le 0`

`\therefore 24x^2+24x+4\le 0`

`6x^2+6x+1\le 0`

`(x+\frac{3+\sqrt3}{6})(x+\frac{3-\sqrt3}{6})\le 0`

`x\in(\frac{-3-\sqrt3}{6}, \frac{-3+\sqrt3}{6})`

Hence, the given function is concave in `(\frac{-3-\sqrt3}{6}, \frac{-3+\sqrt3}{6})`

the given function is convex in `(-\infty, \frac{-3-\sqrt3}{6})\cup (\frac{-3+\sqrt3}{6}, \infty)`