Let f(x) = x^{3} + x^{2}f^{prime}(1) + 2x f^{ | vedclass

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(D) Given $f(x) = x^{3} + x^{2}f^{\prime}(1) + 2x f^{\prime\prime}(2) + f^{\prime\prime\prime}(3)$.First derivative: $f^{\prime}(x) = 3x^{2} + 2x f^{\

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$ \frac{117}{5} $

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(D) Given $f(x) = x^{3} + x^{2}f^{\prime}(1) + 2x f^{\prime\prime}(2) + f^{\prime\prime\prime}(3)$.First derivative: $f^{\prime}(x) = 3x^{2} + 2x f^{\prime}(1) + 2f^{\prime\prime}(2)$.Second derivative: $f^{\prime\prime}(x) = 6x + 2f^{\prime}(1)$.Third derivative: $f^{\prime\prime\prime}(x) = 6$.Now, evaluate the constants:$f^{\prime\prime}(2) = 6(2) + 2f^{\prime}(1) = 12 + 2f^{\prime}(1)$.$f^{\prime\prime\prime}(3) = 6$.Substitute these into the expression for $f^{\prime}(x)$:$f^{\prime}(x) = 3x^{2} + 2x f^{\prime}(1) + 2(12 + 2f^{\prime}(1)) = 3x^{2} + 2x f^{\prime}(1) + 24 + 4f^{\prime}(1)$.Set $x = 1$ to find $f^{\prime}(1)$:$f^{\prime}(1) = 3(1)^{2} + 2(1)f^{\prime}(1) + 24 + 4f^{\prime}(1)$.$f^{\prime}(1) = 3 + 2f^{\prime}(1) + 24 + 4f^{\prime}(1) = 27 + 6f^{\prime}(1)$.$-5f^{\prime}(1) = 27 \implies f^{\prime}(1) = -\frac{27}{5}$.Now substitute $f^{\prime}(1)$ back into $f^{\prime}(x)$:$f^{\prime}(x) = 3x^{2} + 2x(-\frac{27}{5}) + 24 + 4(-\frac{27}{5}) = 3x^{2} - \frac{54}{5}x + 24 - \frac{108}{5} = 3x^{2} - \frac{54}{5}x + \frac{12}{5}$.Finally, calculate $f^{\prime}(5)$:$f^{\prime}(5) = 3(5)^{2} - \frac{54}{5}(5) + \frac{12}{5} = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{105 + 12}{5} = \frac{117}{5}$.

Let $f(x) = x^{3} + x^{2}f^{\prime}(1) + 2x f^{\prime\prime}(2) + f^{\prime\prime\prime}(3)$, where $x \in R$. Then the value of $f^{\prime}(5)$ is:

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