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

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(B) આપેલ છે $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$.પ્રથમ,વિકલન $f'(x)$ શોધો:$f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.$f'(1)$ ની કિંમત શ

Vedclass · Accepted Answer

$\frac{53}{3}$

Vedclass · Answer

(B) આપેલ છે $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$.પ્રથમ,વિકલન $f'(x)$ શોધો:$f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.$f'(1)$ ની કિંમત શોધો:$f'(1) = 30(1)^9 - 56(1)^7 + 30(1)^5 - 63(1)^2 + 6(1) = 30 - 56 + 30 - 63 + 6 = -53$.હવે,લક્ષ $L = \mathop {\lim }\limits_{\alpha 	o 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$ ધ્યાનમાં લો.આ $\frac{0}{0}$ સ્વરૂપ હોવાથી,$L'	ext{Hospital}$ ના નિયમનો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{\alpha 	o 0} \frac{f'(1 - \alpha) \cdot (-1)}{3\alpha^2 + 3}$.$\alpha = 0$ મૂકતા:$L = \frac{f'(1) \cdot (-1)}{3(0)^2 + 3} = \frac{-f'(1)}{3} = \frac{-(-53)}{3} = \frac{53}{3}$.

જો $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$ હોય,તો $\mathop {\lim }\limits_{\alpha \to 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$ ની કિંમત શોધો.

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