Let $f(x)=\begin{cases} x^a\ln x & x>0 \\ 0 & x=0 \end{cases}$. Find the value of $a$ for which Rolle’s theorem is applicable in $[0,1]$

Question

$$f(0) = f(1)=0$$

So according to Rolle’s theorem, there exists a number $c \in (0,1)$ such that $f’(c) =0$

$$f’(x)=x^{a-1} + ax^{a-1} \ln x$$ $$f’(x)=x^{a-1} (1+a\ln x)=0$$ $$a=\frac{-1}{\ln x}$$

How o I find the value of $a$ from here?

Answer 1

You need right continuity at $0$.

Set $e^{-y} :=x$, $y >0$, and consider

$L:= \lim_{y \rightarrow \infty} e^{-ya}(-y)= \lim_{y \rightarrow \infty} \dfrac{-y}{e^{ya}};$

$L=0$ implies $a >0$.

Answer 2

The condition on $a$ is that $a> 0$. We need to show that $f$ is continuous at $x = 0$ with $a > 0$. We have: $f(0) = 0$, and $\displaystyle \lim_{x \to 0^{+}} f(x)= \displaystyle \lim_{x \to 0^{+}} x^a\ln x= \displaystyle \lim_{y \to +\infty} -\dfrac{\ln y}{y^a}= -\displaystyle \lim_{y \to +\infty} \dfrac{\frac{1}{y}}{ay^{a-1}}= -\displaystyle \lim_{y \to +\infty} \dfrac{1}{ay^a}=0$ by L'hospitale rule. And $f$ is also continuous at $0 < x \le 1$,and differentiable on $(0,1)$ for all $a$ as you already had $f'(x) = x^{a-1}(1+a\ln x)$ is defined for any $x > 0$ regardless of what $a$ is. Thus $a > 0$ is the only condition that is needed for the Rolle's Theorem to be applied.