finding the values of a and b(real numbers) that makes the function differentiable at any point of its domain

Question

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?

Answer 1

Correct your continuity condition: it should be $ae+b=1,$ not $a+b=e.$

As for the differentiability condition, you can proceed for the left differential like you did for the right one, but using that now, $f(e)=1:$ $$\lim_{h \to 0^{-}} \frac{f(e+h)-f(e)}h=\lim_{h \to 0^{-}} \frac{\ln^4(e+h)-1}h=\dots$$ However, a simpler method is to differentiate as usual at $e$ the two (now both continuous) functions respectively defined on $(0,e]$ and on $[e,+\infty),$ which are both restrictions of smooth functions on $(0,+\infty).$ You thus get for instance: $$f'_{\rm left}(e)=4\ln^3(e)\frac1e=\frac4e.$$ Beware to compute it directly like this, and not as the limit of $f'(x)$ as $x\to e^-,$ which would be true also but would require some additional argument.

Conclusion: the two conditions are $$ae+b=1,\quad\frac4e=a$$ hence the solution is $$a=\frac4e,\quad b=-3.$$

Answer 2

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 $b=-3$.