I'm given this function:
$$
f(x) = \begin{cases}
(lnx)^4 & 0<x< e \\
ax+b & x≥e
\end{cases}
$$
and I'm asked for which real values of 'a' and 'b' the function is differentiable at any point
I understand that for every x≠1 the function is differentiable.
and now I'm trying the find the values of 'a' and 'b' that make the function differentiable at x=e
this is what I did so far:
first I demanded that the function will be Continuous at x=e, meaning I demand that
$\lim\limits_{x \to e^{-}} f(x)
= \lim\limits_{x \to e^{+}} f(x) = f(e)$
and I got the equation:
$ae+b = e$
next, I demanded that the function will be differentiable at x = e,
I demanded that
$\lim\limits_{h \to 0^{-}} \frac{f(e+h)-f(e)}{h}
= \lim\limits_{h \to 0^{+}} \frac{f(e+h)-f(e)}{h}$
so far I got:
$\lim\limits_{h \to 0^{+}} \frac{f(e+h)-f(e)}{h} = \lim\limits_{h \to 0^{+}} \frac{(a(e+h)+b) – (ae+b)}{h} = \lim\limits_{h \to 0^{+}} \frac{(ae+ah+b) – (ae+b)}{h} = \lim\limits_{h \to 0^{+}} \frac{ah}{h} = \lim\limits_{h \to 0^{+}} a = a$
but when I tried to calculate $\lim\limits_{h \to 0^{-}} \frac{f(e+h)-f(e)}{h}$ I got stuck, so far I got
$\lim\limits_{h \to 0^{-}} \frac{f(e+h)-f(e)}{h} = \lim\limits_{h \to 0^{-}} \frac{(ln(e+h))^4 – (ae+b)}{h} = ???$
now I don't know how to solve this, I'll be glad of any help you can give me,
should I calculate the derivative function at $0<x<e$ and at $x>e$ using the chain rule and then demand that $f'(1+) = f'(1-) $
or should I stick to the limit definition of the derivative?
Best Answer
The First equation is wrong, for continuity you need that $f(e^-) =f(e^+) $, which implies that $ae+b=1$. Now for $$\lim_{t\to 0}\frac{(\ln(e+h))^4-ae-b}{et},$$ use that $ae+b=1$, and $\ln(e+h) =\ln(e(1+h/e))=\ln e+\ln(1+t)\sim 1+t, $ where $t=h/e$. Using that you get the following limit $$\lim_{t\to 0}\frac{(1+t)^4-ae-b}{et}=\lim_{t\to 0}\frac{(1+t)^4-1}{et}=4/e,$$ so we get from continuity that $ae+b=1$ and from differentiability that $a=4/e$, so $b=-3$.