If f(x) = frac{1}{x^2} int_3^x (2t - 3f'(t)) | vedclass

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(C) Given that, $f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$. Multiplying both sides by $x^2$, we get $x^2 f(x) = \int_3^x (2t - 3f'(t)) dt$. Diff

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$\frac{1}{2}$

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(C) Given that, $f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$. Multiplying both sides by $x^2$, we get $x^2 f(x) = \int_3^x (2t - 3f'(t)) dt$. Differentiating both sides with respect to $x$ using the product rule on the left and the Fundamental Theorem of Calculus on the right: $x^2 f'(x) + 2x f(x) = 2x - 3f'(x)$. Rearranging the terms: $f'(x)(x^2 + 3) = 2x - 2x f(x)$. At $x = 3$, note that $f(3) = \frac{1}{3^2} \int_3^3 (2t - 3f'(t)) dt = 0$. Substituting $x = 3$ into the differentiated equation: $f'(3)(3^2 + 3) = 2(3) - 2(3) f(3)$. $f'(3)(9 + 3) = 6 - 6(0)$. $12 f'(3) = 6$. $f'(3) = \frac{6}{12} = \frac{1}{2}$.

If $f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$, then $f'(3)$ is equal to

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