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Update GBranchQ6AU15.tex #960

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29 changes: 1 addition & 28 deletions meanValueTheorem/exercises/GBranchQ6AU15.tex
Original file line number Diff line number Diff line change
Expand Up @@ -11,34 +11,7 @@

Let $f(x) = 20 + 8x^2 - x^4$.\\
Find the global maximum and global minimum values of $f$ on the closed interval $[-3,3]$ and find the x-coordinates of the points where they are attained.
\begin{hint}
First, we have to compute $f'(x)$.

$f'(x) = \answer{16x-4x^3}=-4x(\answer{x^2-4})=-4x(x-\answer{2})(x+\answer{2})$.

Complete the statement below.

The x-coordinates of all critical points of $f$ (from left to right) are $a=\answer{-2}$, $b=\answer{0}$, and $c=\answer{2}$.
\end{hint}
\begin{hint}
Now, we evaluate the function $f$ at the end points and at the critical points.
$$
f(-3) = \answer{11}
$$
$$
f(3) = \answer{11}
$$
$$
f(a) = \answer{36}
$$
$$
f(b) = \answer{20}
$$
$$
f(c) = \answer{36}
$$
Now we compare these values and answer the question.
\end{hint}

The global maximum value of $f$ on the interval $[-3,3]$ is $\answer{36}$ and it is attained at $x=\answer{-2}$ and at $x=\answer{2}$ .

The global minimum value of $f$ on the interval $[-3,3]$ is $\answer{11}$ and it is attained at $x=\answer{-3}$ and at $x=\answer{3}$ .
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