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

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Answer 1 · 2016-12-03T07:21:28

The function is concave.

$f (x) = - x^{2} + 4 x + 2$ $f(x)=-x^2+4x+2$

$f'(x)=-2x+4$

For critical points, $f'(x)=0$

$-2x+4=0$

$x=2$

Let's do a sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $2$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $f'(x)$ $color(white)(aaaa)$ $+$ $color(white)(aa)$ $0$ $color(white)(aaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaa)$ $uarr$ $color(white)(aa)$ $6$ $color(white)(aaa)$ $darr$

So $f(x)$ is concave between $-oo$ and $+oo$

graph{-x^2+4x+2 [-16.02, 16.02, -8.01, 8.01]}

For what values of x is f(x)=-x^2+4x+2f(x)=−x2+4x+2f(x)=-x^2+4x+2 concave or convex?