If f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^ | vedclass

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(A) Given $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$. First,find the derivative $f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$. Evaluating at $x

Vedclass · Accepted Answer

$\frac{53}{3}$

Vedclass · Answer

(A) Given $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$. First,find the derivative $f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$. Evaluating at $x = 1$,$f'(1) = 30 - 56 + 30 - 63 + 6 = -53$. Now,consider the limit $L = \lim_{\alpha ightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha}$. Rewrite the expression as $L = \lim_{\alpha ightarrow 0} \left( \frac{f(1-\alpha) - f(1)}{-\alpha} ight) 	imes \left( \frac{-\alpha}{\alpha^3 + 3\alpha} ight)$. Since $\lim_{\alpha ightarrow 0} \frac{f(1-\alpha) - f(1)}{-\alpha} = f'(1) = -53$ and $\lim_{\alpha ightarrow 0} \frac{-\alpha}{\alpha(\alpha^2 + 3)} = \lim_{\alpha ightarrow 0} \frac{-1}{\alpha^2 + 3} = -\frac{1}{3}$. Therefore,$L = (-53) 	imes (-1/3) = \frac{53}{3}$.

If $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$,then $\lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha} = $

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