Questions: You were asked to find all absolute extrema for the function (f(x)=4 x^3+30 x^2+48 x) on the interval ([-9,8]). Now that we know that (f'(x)=12 x^2+60 x+48), we can find the critical points. Find the values of where (f'(x)=0) and the values of (x) where (f'(x)) does not exist. Discard any solution that is outside ([-9,8]). Separate multiple answers with a comma where necessary.

You were asked to find all absolute extrema for the function (f(x)=4 x^3+30 x^2+48 x) on the interval ([-9,8]).

Now that we know that (f'(x)=12 x^2+60 x+48), we can find the critical points. Find the values of where (f'(x)=0) and the values of (x) where (f'(x)) does not exist. Discard any solution that is outside ([-9,8]). Separate multiple answers with a comma where necessary.

Transcript text: You were asked to find all absolute extrema for the function $f(x)=4 x^{3}+30 x^{2}+48 x$ on the inter $[-9,8]$. Now that we know that $f^{\prime}(x)=12 x^{2}+60 x+48$, we can find the critical points. Find the values of where $f^{\prime}(x)=0$ and the values of $x$ where $f^{\prime}(x)$ does not exist. Discard any solution that is outside $[-9,8]$. Separate multiple answers with a comma where necessary.

Solution

Find the critical points by solving $f'(x) = 0$ and identifying where $f'(x)$ does not exist within the interval $[-9, 8]$.

Find where $f'(x) = 0$.

We have $f'(x) = 12x^2 + 60x + 48$. Setting this equal to 0 gives: $12x^2 + 60x + 48 = 0$ $12(x^2 + 5x + 4) = 0$ $x^2 + 5x + 4 = 0$ $(x + 1)(x + 4) = 0$ $x = -1$ or $x = -4$

Find where $f'(x)$ does not exist.

Since $f'(x)$ is a polynomial, it exists for all real numbers $x$.

Determine which critical points are within the interval $[-9, 8]$.

Both $x = -1$ and $x = -4$ are within the interval $[-9, 8]$.

$\boxed{-4, -1}$

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