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

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Answer 1 · 2018-07-08T04:04:42

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

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

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

Find the first derivative by using the power rule : $(u^n)' = n u^(n-1)u'$

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

$f'(x) = 2(3-x)(-1) - 1$

$f'(x) = -(6-2x) - 1$

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

$f'(x) = 2x - 7 = 0$

Critical value: $" "2x = 7; " "x = 7/2$

Find the first derivative by using distribution :

$f(x) = (3-x)(3-x) - x$

$f(x) = 9 - 6x + x^2 - x$

$f(x) = x^2 -7x + 9$

$f'(x) = 2x - 7 = 0$

Critical value: $" "2x = 7; " "x = 7/2$

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