From b4e753b55d50838db5f3599f074569a7bc80806f Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Mon, 7 Sep 2015 12:56:20 +0200
Subject: fix 1-1b

---
 assignment1.tex | 13 +++++++------
 1 file changed, 7 insertions(+), 6 deletions(-)

(limited to 'assignment1.tex')

diff --git a/assignment1.tex b/assignment1.tex
index eeeb3a6..ad96472 100644
--- a/assignment1.tex
+++ b/assignment1.tex
@@ -39,13 +39,14 @@
                 \end{align*}
 
                 This gives $x=\sqrt1=1$ and $x=-\sqrt1=-1$ as solutions, yielding $-1,0,1$ as possible values for x.
-            \item \begin{align*}
-                    x - x^3 &> 0\\
-                    x &> x^3 \qquad\text{We see $x=0$ is no solution, so we can assume $x\neq0$.}\\
-                    1 &> x^2
-                \end{align*}
+            \item First, we see that $x=0$ is no solution. We then make a case distinction:
+
+                \begin{itemize}
+                    \item Case $x>0$. It follows that $x > x^3$ and $1 > x^2$, hence $x\in(0,1)$.
+                    \item Case $x<0$. It follows that $x > x^3$ and $1 < x^2$, hence $x\in(-\infty,-1)$.
+                \end{itemize}
 
-                We know $1=x^2$ gives $x=-1$ or $x=1$, so this inequality gives intuitively $-1>x$ or $1 > x > 0$. This gives $x\in(-\infty,-1)\cup(0,1)$.
+                The values of $x$ for which $f(x)>0$ are then $(\infty,-1)\cup(0,1)$.
         \end{enumerate}
 
     \item For particular values of $a,b$ the function $y=a+(b-a)x$ is a linear function, hence it has extremes at $x=0$ and $x=1$. This gives the extremes $y=a+(b-a)\cdot0=a$ and $y=a+(b-a)\cdot1=b$. Therefore, $y$ runs through $(a,b)$. Note that strictly we calculated the extremes on $[0,1]$ rather than $(0,1)$ but that this does not matter for a linear function and that we corrected this by giving the range $(a,b)$ rather than $[a,b]$.
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