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

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Answer 1 · 2018-07-27T22:17:40

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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 $(-\infty, \frac{-3-\sqrt3}{6})\cup (\frac{-3+\sqrt3}{6}, \infty)$

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