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Diffstat (limited to 'assignment1.tex')
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1 files changed, 7 insertions, 6 deletions
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 $[a,b]$. |