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 $x=\answer{3}$ .