Animated Solution for Mathematics - Differentiation: If $f(x)=x^{\alpha }\log x$ and $f(0)=0$, then the value of $\alpha$ for which Rolle's theorem can be applied in $[0,1]$ is

Rolle's Theorem requires a function $f(x)$ to satisfy three fundamental conditions on an interval $[a,b]$: 1. $f(x)$ must be continuous on $[a,b]$. 2. $f(x)$ must be differentiable on $(a,b)$. 3. The endpoints must satisfy $f(a)=f(b)$.

We are investigating the function $f(x)=x^{\alpha }\log x$ on the interval $[0,1]$.

First, we examine the boundaries of the interval. We are given $f(0)=0$ by definition.

Now, we evaluate the function at $x=1$:

Since $f(0)=f(1)=0$, the third condition of Rolle's Theorem is satisfied for any finite value of $\alpha$.

The true test of our function occurs at the origin, $x=0$. For $f(x)$ to be continuous on $[0,1]$, the limit as $x$ approaches $0$ from the right must equal $f(0)$.

We must evaluate:

If $\alpha \le 0$, the function fails to be continuous. For $\alpha =0$, the function becomes $\log x$, which approaches $-\infty$ as $x\to 0^{+}$. For $\alpha <0$, the term $x^{\alpha }$ approaches $\infty$, and the product with $\log x$ drives the function to $-\infty$.

To ensure continuity, we must have $\alpha >0$. With $\alpha >0$, we face the indeterminate form $0\cdot (-\infty )$. We rewrite the limit to apply L'Hopital's Rule:

Applying L'Hopital's Rule by differentiating the numerator and denominator:

Since $\alpha >0$, as $x\to 0^{+}$, the expression $x^{\alpha }$ vanishes to $0$. Thus, the limit is $0$, which matches $f(0)$, confirming continuity.

Finally, we examine the derivative $f^{\prime }(x)$:

For this derivative to exist on the open interval $(0,1)$, the term $x^{\alpha -1}$ must be well-defined. This requires the exponent to be non-negative, or more specifically, that the function does not blow up at the boundary $x=0$.

This condition is satisfied when $\alpha -1>-1$, which simplifies to $\alpha >0$.

We have arrived at our destination: Rolle's Theorem is applicable to $f(x)=x^{\alpha }\log x$ on $[0,1]$ if and only if $\alpha >0$.

Any value of $\alpha$ greater than zero satisfies the requirements of the theorem.