From 34fcb097b0caac6cb383ccb9b6acb8518fedb8de Mon Sep 17 00:00:00 2001
From: pfatheddin <156558883+pfatheddin@users.noreply.github.com>
Date: Sun, 24 Mar 2024 17:50:49 -0400
Subject: [PATCH] Update GBranchQ6AU15.tex

https://ximera.osu.edu/mooculus/meanValueTheorem/exercises/exerciseList/meanValueTheorem/exercises/GBranchQ6AU15

I took out the hint. The problem before this had the steps in hint and they just need to practice here. The hint makes it longer and they should know the steps.
---
 meanValueTheorem/exercises/GBranchQ6AU15.tex | 29 +-------------------
 1 file changed, 1 insertion(+), 28 deletions(-)

diff --git a/meanValueTheorem/exercises/GBranchQ6AU15.tex b/meanValueTheorem/exercises/GBranchQ6AU15.tex
index ee6e40a78..cd70c0a46 100644
--- a/meanValueTheorem/exercises/GBranchQ6AU15.tex
+++ b/meanValueTheorem/exercises/GBranchQ6AU15.tex
@@ -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 $x=\answer{3}$ .