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If $f(x) = x^{\alpha}\cdot \ln(x)$ and $f(0)=0$. Then the value of $\alpha$ for which Rolles Theorem is applicable, is

$\bf{My\; Try::}$ Rolle,s Theorem is Applicable in $x\in \left[\; 0,1\right]$

$(1)\; $ If function $f(x)$ is Continuous in $\left[\; 0,1\right].$

$(2)\;$ If function $f(x)$ is Differentiable in $(0,1)$

$(3)\;$ And $f(0) = f(1) = 0$

Then There exists at least one value of $x=c$ for which $f^{'}(c) = 0$

Where $x=c\in (0,1)$

Now I did not understand how can i Solve it

Help me

Thanks

Christian Blatter
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juantheron
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1 Answers1

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You need to verify each condition you have mentioned. In particular, you need to check for what $\alpha$ we have $\lim_{x\to 0^+}x^\alpha\ln(x) = 0$

If $\alpha<0$, then $\lim_{x\to 0^+}x^\alpha = +\infty, \lim_{x\to 0^+}\ln(x) = -\infty$.

If $\alpha=0$, then $x^\alpha = 1, \lim_{x\to 0^+}\ln(x) = -\infty$.

If $\alpha>0$, use the fact $\lim_{x\to 0^+}x\ln(x) = 0$ and $x^\alpha\ln(x) = \dfrac{x^\alpha \ln(x^\alpha)}{\alpha}$ to get the limit looked for.

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Petite Etincelle
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