So I have this question here.
Some background. I already took calculus(Single/Multivariable) along with many other math courses so I would know how to do this questions fairly easily. The constraints in this question is what's tripping me up.
For part $i)$, $m=\frac{5}{8}$ and $b=\frac{3}{2}$.
I'm not really sure what part $ii)$ is asking to be fair. I can't use the rules of differentiation so i'm stuck with using first principles, (i.e. the definition).
If I attempt to use the definition that they give me, then:
$f_{-}'(4)=\displaystyle{\lim_{x \to 0}} \frac{f(4+h)-f(4)}{h}$
and
$f_{+}'(4)=\displaystyle{\lim_{h \to 0}} \frac{f(4+h)-f(4)}{h}$
But now like, what to I substitute in? My intuition is telling me for the first limit, I would sub in the $\frac{2}{x+2}$ to get:
$f_{-}'(4)=\displaystyle{\lim_{h \to 0}} \frac{\frac{2}{x+h+2}-\frac{4}{4+2}}{h}$
However, if I attempt to do that for the other limit, I can't evaluate the $f(4)$ term since the limit from the right is undefined.
My second ideas was to use the function in the middle, namely:
$f_{+}'(4)=\displaystyle{\lim_{h \to 0}} \frac{(\frac{5}{8}(x+h)+\frac{3}{2})-(\frac{5}{8}(x)+\frac{3}{2})}{h}$
$f_{+}'(4)=\displaystyle{\lim_{h \to 0}} \frac{(\frac{5}{8}(4+h)+\frac{3}{2})-(\frac{5}{8}(4)+\frac{3}{2})}{h}$
and then proceed but then, I'm not really sure if this technique i'm using is even right. Can someone nudge me in the right direction? Thanks!
Note that your function for $x>4$ is actually $$\frac {x-4}{\sqrt x -2} = \frac {(\sqrt x-2)(\sqrt x+2)}{\sqrt x-2}= \sqrt x+2$$ That should make your computations straight forward.