For what values of x is $f (x) = 3 + \frac{x^{2}}{2}$ $f(x)=3+x^2/2$ concave or convex?

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Answer 1 · 2016-01-01T18:26:42

$f$ $f$ is convex on $(- \infty, + \infty)$ $(-oo,+oo)$ .

$f$ $f$ is convex when $f''>0$ .
$f$ is concave when $f''<0$ .

First, find $f'(x)$ , then $f''(x)$ .

$f(x)=3+1/2x^2$

$f'(x)=x$

$f''(x)=1$

Since $f''(x)$ is always $>0$ , $f$ is convex on $(-oo,+oo)$ .

graph{3+(x^2)/2 [-14.96, 17.08, -1.67, 14.35]}

The answer makes logical since, since $f$ is a parabola that opens up, so it will always be in the convex shape.

For what values of x is f(x)=3+x^2/2f(x)=3+x22f(x)=3+x^2/2 concave or convex?