What are the critical values, if any, of $f(x)= (x^2+6x-7)^2$ ?

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Answer 1 · 2015-11-19T04:03:47

The critical values for $f$ are $-7,-3, "and " 1$

Use the power rule and the chain rule to find:

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

Since, $f(x)$ and $f'(x)$ are polynomials, the critical values for $f$ are the zeros of $f'$

Solve $f'(x) = 2(x^2+6x-7)(2x+6)=0$ by solving

$x^2+6x-7=0$ or $2x+6=0$ . So,

$(x+7)(x-1)=0$ or $x=-3$

$x=-7, 1, "or" -3$

The critical values for $f$ are $-7,-3, "and " 1$

What are the critical values, if any, of f(x)= (x^2+6x-7)^2f(x)= (x^2+6x-7)^2?