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### Answers (1)

Answer:

$c=0 \in(-1,1)$

Hint:

$f(-1)=f(1)$ , so there exists at $c \in(-1,1)$

Given:

$f(x)=\log \left(x^{2}+2\right)-\log3$on$[-1,1]$

Explanation:

We have

$f(x)=\log \left(x^{2}+2\right)-\log3$on$[-1,1]$

We know that logarithmic function is continuous and differentiable on its own domain.

Let’s find the values of the function at an extreme.

$\\ \Rightarrow \quad f(-1)=\log \left((-1)^{2}+2\right)-\log 3 \\\\ \Rightarrow \quad f(-1)=\log (1+2)-\log 3\\\\ \Rightarrow \quad f(-1)=\log 3-\log 3\\\\ \therefore \quad f(-1)=0\\\\ \Rightarrow \quad f(1)=\log \left(1^{2}+2\right)-\log 3\\\\ \Rightarrow \quad f(1)=\log (1+2)-\log 3\\\\ \Rightarrow \quad f(1)=\log 3-\log 3\\\\ \therefore \quad f(1)=0$

We have  $f(-1)=f(1)$ , so there exists at $c \in(-1,1)$, such that$f'(c) = 0$

Let’s find the derivative of f(x)

$f^{\prime}(x)=\frac{d\left(\log \left(x^{2}+2\right)-\log 3\right)}{d x}$

$\\ \Rightarrow \quad f^{\prime}(x)=\frac{1}{x^{2}+2} \frac{d\left(x^{2}+2\right)}{d x}=0 \ \ \ \ \ \ \left[\because \frac{d(\log x)}{d x}=\frac{1}{x}\right] \\\\ \Rightarrow \quad f^{\prime}(x)=\frac{2 x}{x^{2}+2}$

We have $f'(c) = 0$

$\\ \Rightarrow \quad \frac{2 c}{c^{2}+2}=0\\\\ \Rightarrow 2c = 0 \\\\ \Rightarrow \quad c=0 \in(-1,1)$

Hence, Rolle’s Theorem is verified.

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