Question
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=3+(\text{x}-2)^{\frac{2}{3}}\text{ on }[1,3]$
$\text{f}(\text{x})=3+(\text{x}-2)^{\frac{2}{3}}\text{ on }[1,3]$
✓
Answer
The given function is $\text{f}(\text{x})=3+(\text{x}-2)^{\frac{2}{3}}$
Defferentiating with respect to x, we get
$\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{2}{3}-1}$
$\Rightarrow\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{-1}{3}}$
$\Rightarrow\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{1}{3}}$
Clearly, we observe that for $\text{x}=2\in[1,3],\text{f}'(\text{x})$ does not exist.
Therefore, f(x) is not derivable on [1, 3].
Hence, Rolle's theorem is not applicable for the given function.
Defferentiating with respect to x, we get
$\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{2}{3}-1}$
$\Rightarrow\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{-1}{3}}$
$\Rightarrow\text{f}'(\text{x})=\frac{2}{3}(\text{x}-2)^{\frac{1}{3}}$
Clearly, we observe that for $\text{x}=2\in[1,3],\text{f}'(\text{x})$ does not exist.
Therefore, f(x) is not derivable on [1, 3].
Hence, Rolle's theorem is not applicable for the given function.
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