What are the critical values, if any, of $f(x) = -3x^2+x-1$ ?

Question

1636 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-22T06:08:11

The critical value is a maximum at $(1/6,-11/12)$

We must differentiate and do a sign chart

We use

$(x^n)'=nx^(n-1)$

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

$f'(x)=-6x+1$

$f'(x)=0$ , when $-6x+1=0$ , $=>$ , $x=1/6$

$f(1/6)=-3/36+1/6-1=-11/12$

We can do our sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaa)$ $-oo$ $color(white)(aaaaa)$ $1/6$ $color(white)(aaaaa)$ $+oo$

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

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $uarr$ $color(white)(aa)$ $-11/12$ $color(white)(aaa)$ $darr$

The critical value is a maximum at $(1/6,-11/12)$

What are the critical values, if any, of f(x) = -3x^2+x-1f(x) = -3x^2+x-1?