What are the critical values of f(x)=(3-x)^2-x?

Jul 8, 2018

$x = \frac{7}{2}$

Explanation:

Given: $f \left(x\right) = {\left(3 - x\right)}^{2} - x$

You find critical values by setting $f ' \left(x\right) = 0$ and solving for $x$.

Find the first derivative by using the power rule : $\left({u}^{n}\right) ' = n {u}^{n - 1} u '$

Let u = 3-x; " "u' = -1; " " n = 2

$f ' \left(x\right) = 2 \left(3 - x\right) \left(- 1\right) - 1$

$f ' \left(x\right) = - \left(6 - 2 x\right) - 1$

$f ' \left(x\right) = - 6 + 2 x - 1 = 2 x - 7$

$f ' \left(x\right) = 2 x - 7 = 0$

Critical value: $\text{ "2x = 7; " } x = \frac{7}{2}$

Find the first derivative by using distribution :

$f \left(x\right) = \left(3 - x\right) \left(3 - x\right) - x$

$f \left(x\right) = 9 - 6 x + {x}^{2} - x$

$f \left(x\right) = {x}^{2} - 7 x + 9$

$f ' \left(x\right) = 2 x - 7 = 0$

Critical value: $\text{ "2x = 7; " } x = \frac{7}{2}$