From f2022fbfb518969cff03b7cadd313cec57b345fc Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Fri, 18 Sep 2015 16:54:41 +0200
Subject: Finish assignment 3

---
 assignment3.tex | 42 ++++++++++++++++++++++++++++++++++++++----
 1 file changed, 38 insertions(+), 4 deletions(-)

(limited to 'assignment3.tex')

diff --git a/assignment3.tex b/assignment3.tex
index 8428b35..cb6165a 100644
--- a/assignment3.tex
+++ b/assignment3.tex
@@ -161,15 +161,49 @@
                 
             \item We take $\{(x,f(x)) \mid f'(x)=0\} = \{(2,f(2)),(-6,f(-6))\} = \{(2,0),(-6,-16)\}$.
                 
-            \item At $x=2$ we have $f''(x)=\frac12>0$, so $(2,0)$ is a local minimum. At $x=-6$ we have $f''(x)=-\frac12<0$, so $(-6,-16)$ is a local maximum.
+            \item At $x=2$ we have $f''(x)=\frac12>0$, so $(2,0)$ is a local minimum. 
                 
-            \item %todo
+                At $x=-6$ we have $f''(x)=-\frac12<0$, so $(-6,-16)$ is a local maximum.
                 
-            \item %todo
+            \item From $f''(x) > 0$ follows $(x+2)^3 > 0$ and so $x > -2$. Therefore, $f$ is convex on $(-2,\infty)$.
+
+                From $f''(x) < 0$ follows $(x+2)^3 < 0$ and so $x < -2$. Therefore, $f$ is concave on $(-\infty,-2)$.
+
+                On $x=-2$ we could have had a point of inflection (it's the only $x\in\mathbb R$ on which $f$ is neither convex nor concave), however, $-2$ is not in the domain of $f$. Therefore, this function does not have a point of inflection.
+                
+            \item We have 
+                \begin{align*}
+                    f(x) - (x-6) &= \frac{x^2-4x+4}{x+2} - \frac{x^2 - 4x + 36}{x+2}\\
+                                 &= -\frac{32}{x+2}.
+                \end{align*}
+                
+                Then obviously both limits mentioned in the exercise are equal to zero. Therefore, $y=x-6$ is a slant asymptote of $f$.
                 
         \end{enumerate}
 
-    \item %todo
+    \item \begin{enumerate}
+            \item \begin{align*}
+                    f'(x)   &= \frac1{\cos(\ln(\cos x))}\cdot\cos(\ln(\cos x))'\\
+                            &= \frac1{\cos(\ln(\cos x))}\cdot -\sin(\ln(\cos x))\cdot (\ln(\cos x))'\\
+                            &= -\frac{\sin(\ln(\cos x))}{\cos(\ln(\cos x))}\cdot \frac1{\cos x}\cdot (\cos x)'\\
+                            &= -\frac{\sin(\ln(\cos x))}{\cos(\ln(\cos x))\cdot\cos x}\cdot -\sin x\\
+                            &= \frac{\sin(\ln(\cos x))\cdot\sin x}{\cos(\ln(\cos x))\cdot\cos x}\\
+                            &= \tan(\ln(\cos x))\cdot\tan x.
+                \end{align*}
+                
+            \item For example $f(x) = -\frac12\ln(\cos(2x))$ works:
+                \begin{align*}
+                    f'(x)   &= -\tfrac12\cdot\frac1{\cos(2x)}\cdot(\cos(2x))'\\
+                            &= -\tfrac12\cdot\frac1{\cos(2x)}\cdot-\sin(2x)\cdot2\\
+                            &= \frac{\sin(2x)}{\cos(2x)}\\
+                            &= \tan(2x).
+                \end{align*}
+                
+            \item For example take $f_1(x) = -\frac14\cos(2x)$, then we have the derivative $$f_1'(x) = -\frac14\cdot-\sin(2x)\cdot2 = \frac12\sin(x+x) = \frac12(\sin x\cos x +\sin x\cos x) = \sin x\cos x.$$
+
+                Now we define $f_i(x) = -\frac14\cos(2x) + i - 1$ for all $i \in\mathbb R$, which gives us infinitely many functions with the same derivative.
+                
+        \end{enumerate}
 
 \end{enumerate}
 
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