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

1 Answer

Answer:

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)#