Let f(x) be a polynomial function such that f | vedclass

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(A) Given $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$. Since the limit $\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$ exists,we must have $f(1)=0$.

Vedclass · Accepted Answer

$-15$

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(A) Given $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$. Since the limit $\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$ exists,we must have $f(1)=0$. If $f(1)=0$,then the limit is equal to $f^{\prime}(1)$. Substituting $x=1$ in the given equation: $f(1)+f^{\prime}(1)+f^{\prime \prime}(1) = 1^{5}+64 = 65$. Since $f(1)=0$,we have $f^{\prime}(1)+f^{\prime \prime}(1) = 65$. Differentiating the given equation: $f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x) = 5x^{4}$. At $x=1$,$f^{\prime}(1)+f^{\prime \prime}(1)+f^{\prime \prime \prime}(1) = 5$. Since $f^{\prime}(1)+f^{\prime \prime}(1) = 65$,we get $65+f^{\prime \prime \prime}(1) = 5$,so $f^{\prime \prime \prime}(1) = -60$. Differentiating again: $f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+f^{(4)}(x) = 20x^{3}$. At $x=1$,$f^{\prime \prime}(1)+f^{\prime \prime \prime}(1)+f^{(4)}(1) = 20$. Differentiating again: $f^{\prime \prime \prime}(x)+f^{(4)}(x)+f^{(5)}(x) = 60x^{2}$. At $x=1$,$f^{\prime \prime \prime}(1)+f^{(4)}(1)+f^{(5)}(1) = 60$. Since $f(x)$ is a polynomial of degree $5$,let $f(x) = ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+g$. Then $f^{(5)}(x) = 120a$. From $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$,the leading coefficient $a=1$. Thus $f^{(5)}(x) = 120$. Using $f^{\prime \prime \prime}(1)+f^{(4)}(1)+120 = 60$,we get $f^{\prime \prime \prime}(1)+f^{(4)}(1) = -60$. From $f^{\prime \prime}(1)+f^{\prime \prime \prime}(1)+f^{(4)}(1) = 20$,we get $f^{\prime \prime}(1) + (-60) = 20$,so $f^{\prime \prime}(1) = 80$. Finally,$f^{\prime}(1) = 65 - f^{\prime \prime}(1) = 65 - 80 = -15$.

Let $f(x)$ be a polynomial function such that $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$. Then,the value of $\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$ is:

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