Solution assignment 06 Absolute value |x|

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Assignment 6

Draw the graph of $f(x)=|x^2-1|+|x|$

Solution

We apply the definition of the absolute value to both terms.

$|x^2-1|=x^2-1$ if $x^2-1\geq0$ , so if $x\leq-1$ or $x\geq1$ (see also assignment 4). Call this interval A.

$|x^2-1|=-x^2+1$ if $-1<x<1$ . Call this interval B.

$|x|=x$ if $x\geq0$ . Call this interval C.

$|x|=-x$ if $x<0$ . Call this interval D.

To draw the graph we distinguish the following intervals.

$x\leq-1$ , this is the union of the intervals A and D, briefly AD.

$-1<x<0$ , this is the union of the intervals B and D, briefly BD.

$0\leq{x}<1$ , this is the union of the intervals B and C, briefly BC.

$x\geq1$ , this is the union of the intervals A and C, briefly AC.

For AD we have: $f(x)=x^2-1+(-x)=x^2-x-1$

For BD we have: $f(x)=-x^2+1+(-x)=-x^2-x+1$

For BC we have: $f(x)=-x^2+1+x=-x^2+x+1$

For AC we have: $f(x)=x^2-1+x=x^2+x-1$

The resulting graph is built up by portions of four different parabolas. The first part intercepts $(0,-1)$ and has a minimum for $x=\displaystyle\frac{1}{2}$ . Calculate the same characteristics for the other parabolas and explain them based on the chart. Note that the graph fits the given function.

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