$f(x) = \begin{cases} x^a\log x, & \text{if $x \neq 0$,} \\[2ex] 0, & \text{if $x=0$. } \end{cases} $
What should be the value of $a$ so that f satisfies Rolle's theorem in [0,1] ??
What I have done is from $f'(x)=0,$ I get $x=e^{-\frac 1a}\in (0,1) \Rightarrow a>0$ and from $f^"(e^{-\frac 1a})>0$ whenever $a>0$. So is this the exact condition or I have missed any condition and $a$ can be refined more?
On
$f$ should be continuous in $[0,1]$ and differentiable in $(0,1)$, and $f(0)=f(1)$.
Your $f$ is differentiable in $(0,1)$, for all $a\in\mathbb R$. But it is continuous only for $a>0$. And, also for $a>0$, we have $f(0)=f(1)=0$.
Hence Rolle's Theorem applies if and only if $a>0$.
On
$f(0)=0$ and $f(1)=0$. $f$ ishould be continuous because $\lim_{x\to 0} f(x)=\lim_{x\to 0} \frac {\log(x)}{\frac {1}{x}}=\lim_{x\to 0} \frac {\frac {1}{x}}{\frac {-ax^{a-1}}{x^{2a}}}=\lim_{x\to 0} \frac {x^a}{-a}=0$ iff $a>0$ from Hospital's rule or else for $a<0$ then $\lim_{x\to 0} f(x)=+\infty$.
On
Your answer is correct, but I wouldn't motivate my answer as you did.
To satisfy Rolle's Theorem you need the function to be continuous (1) in [0,1], to be derivable (2) in ]0, 1[, and you need (3) $f(0) = f(1)$.
1) Obviously the function is continuous when $x \neq 0$, therefore you just have to see when $lim_{x \to 0} \ x^a \log x = 0 $ (answer: $a > 0$)
2) derivate the function and get $f'(x) = (a\log x +1) x^a$. That's ok for ]0,1[, f'(x) is continuous therefore f(x) was derivable.
3) you get $f(0) = 0$ by definition, and $f(1) = 1^a \log 1 = 0$
Therefore you function satisfies Rolle's Theorem for any $a > 0$, as you said.
P.S. Why did you calculate the 2nd derivative?
You need that f is continous on $[0,1]$ (apart from other prerequisites which are obviously fulfilled, that is), i.e. $\lim_{x\rightarrow 0} f(x) = 0$. By well known properties of $\log$ this is true for your $f$ iff $a>0$.