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

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

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$\frac{53}{3}$

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(B) Given $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$.First,find the derivative $f'(x)$:$f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.Evaluate $f'(1)$:$f'(1) = 30(1)^9 - 56(1)^7 + 30(1)^5 - 63(1)^2 + 6(1) = 30 - 56 + 30 - 63 + 6 = -53$.Now,consider the limit $L = \mathop {\lim }\limits_{\alpha 	o 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$.Since this is a $\frac{0}{0}$ form,apply $L'	ext{Hospital's rule}$:$L = \mathop {\lim }\limits_{\alpha 	o 0} \frac{\frac{d}{d\alpha}(f(1 - \alpha) - f(1))}{\frac{d}{d\alpha}(\alpha^3 + 3\alpha)} = \mathop {\lim }\limits_{\alpha 	o 0} \frac{f'(1 - \alpha) \cdot (-1)}{3\alpha^2 + 3}$.Substitute $\alpha = 0$:$L = \frac{f'(1) \cdot (-1)}{3(0)^2 + 3} = \frac{-f'(1)}{3} = \frac{-(-53)}{3} = \frac{53}{3}$.

If $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$,then $\mathop {\lim }\limits_{\alpha \to 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$ is

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