I am having trouble with the following question.
If we fix $a\in[-1,1]$ and define $h_a(x)=\frac{1}{2}(|x-a|+(x-a)))$
I am trying to sketch the graph of $h_{\frac{1}{2}}$ on $[-1,1]$
What does the $\frac{1}{2}$ refer to in $h_\frac{1}{2}$. Do I need to plug that in for $a$ in the equation $h_a(x)=\frac{1}{2}(|x-a|+(x-a)))$?
I have provided a link to a wolfram graph however I am not sure if this is correct or not,
https://www.wolframalpha.com/input/?i=1%2F2(%7Cx-a%7C%2B(x-a))+where+a%3D1%2F2
Whenever a function has a modulus sign in it, you must consider separately the cases when the contents of the modulus is negative, and when it is positive. e.g. $y=|x|$ can be defined as $y=x , x\geq 0$ and $y=-x , x<0$
We have $f_{a}(x) = \frac{1}{2}[ |x-a|+(x-a)]$
Then $f_{\frac{1}{2}}(x) = \frac{1}{2}[ |x-\frac{1}{2}|+(x-\frac{1}{2})]$
We have two cases to consider:
Case $(1)$: $$x-\frac{1}{2} \geq 0 \implies x\geq \frac{1}{2}$$
Case $(2)$: $$x<\frac{1}{2}$$
Note that in general:
$|x| = x$ for $x \geq 0$ and $|x| = -x$ for $x<0$
So firstly consider case $(1)$:
$|x-\frac{1}{2}| = x-\frac{1}{2} \implies f_{\frac{1}{2}}(x) = \frac{1}{2}[ x-\frac{1}{2}+x-\frac{1}{2}] =x-\frac{1}{2}$
Then consider case $(2)$:
$|x-\frac{1}{2}| = \frac{1}{2} - x \implies f_{\frac{1}{2}}(x) = \frac{1}{2}[ -x+\frac{1}{2}+x-\frac{1}{2}] = 0$
So for $x\geq \frac{1}{2}, f_{\frac{1}{2}}(x) = x-\frac{1}{2}$
and for $x<\frac{1}{2}, f_{\frac{1}{2}}(x)=0$ which should be simple to sketch