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

Dec 22, 2016

The critical value is a maximum at $\left(\frac{1}{6} , - \frac{11}{12}\right)$

Explanation:

We must differentiate and do a sign chart

We use

$\left({x}^{n}\right) ' = n {x}^{n - 1}$

$f \left(x\right) = - 3 {x}^{2} + x - 1$

$f ' \left(x\right) = - 6 x + 1$

$f ' \left(x\right) = 0$, when $- 6 x + 1 = 0$, $\implies$, $x = \frac{1}{6}$

$f \left(\frac{1}{6}\right) = - \frac{3}{36} + \frac{1}{6} - 1 = - \frac{11}{12}$

We can do our sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$\frac{1}{6}$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$\uparrow$$\textcolor{w h i t e}{a a}$$- \frac{11}{12}$$\textcolor{w h i t e}{a a a}$$\downarrow$

The critical value is a maximum at $\left(\frac{1}{6} , - \frac{11}{12}\right)$