If Rolle's theorem is applicable for the func | vedclass

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(D) Given $f(x) = \log(x^2 + a) - \log(x)$.Since Rolle's theorem is applicable, $f(3) = f(4)$.$\log(9+a) - \log(3) = \log(16+a) - \log(4) \implies \fr

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$\frac{1}{12}$

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(D) Given $f(x) = \log(x^2 + a) - \log(x)$.Since Rolle's theorem is applicable, $f(3) = f(4)$.$\log(9+a) - \log(3) = \log(16+a) - \log(4) \implies \frac{9+a}{3} = \frac{16+a}{4}$.$36 + 4a = 48 + 3a \implies a = 12$.Thus, $f(x) = \log(x^2 + 12) - \log(x)$.$f'(x) = \frac{2x}{x^2+12} - \frac{1}{x} = \frac{2x^2 - (x^2+12)}{x(x^2+12)} = \frac{x^2-12}{x(x^2+12)}$.Given $f'(c) = 0$, we have $c^2 - 12 = 0 \implies c^2 = 12$.$f''(x) = \frac{d}{dx} \left( \frac{x^2-12}{x^3+12x} \right) = \frac{(2x)(x^3+12x) - (x^2-12)(3x^2+12)}{(x^3+12x)^2}$.Substitute $c^2=12$: $f''(c) = \frac{(2c)(c^3+12c) - (0)(3c^2+12)}{(c^3+12c)^2} = \frac{2c^4+24c^2}{(c^3+12c)^2} = \frac{2(12)^2 + 24(12)}{(12c+12c)^2} = \frac{288+288}{(24c)^2} = \frac{576}{576c^2} = \frac{1}{c^2} = \frac{1}{12}$.

If Rolle's theorem is applicable for the function $f(x) = \log \left( \frac{x^2 + a}{x} \right)$ on $[3, 4]$ with $c \in (3, 4)$ such that $f'(c) = 0$, then the value of $f''(c)$ is

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