The sum of the terms of an infinitely decreas | vedclass

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(C) Given $f(x) = x^3 + 3x - 9$ on $[-2, 3]$.First,find the derivative $f'(x) = 3x^2 + 3$.Since $f'(x) > 0$ for all $x$,the function is strictly incre

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$\frac{2}{3}$

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(C) Given $f(x) = x^3 + 3x - 9$ on $[-2, 3]$.First,find the derivative $f'(x) = 3x^2 + 3$.Since $f'(x) > 0$ for all $x$,the function is strictly increasing.Thus,the maximum value on $[-2, 3]$ occurs at $x = 3$.$f(3) = 3^3 + 3(3) - 9 = 27 + 9 - 9 = 27$.Let the first term be $a$ and common ratio be $r$. The sum of an infinite $G$.$P$. is $S = \frac{a}{1-r} = 27$ (Equation $i$).The difference between the first and second term is $a - ar = f'(0)$.$f'(0) = 3(0)^2 + 3 = 3$.So,$a(1-r) = 3$ (Equation $ii$).From $(i)$,$a = 27(1-r)$. Substitute into $(ii)$:$27(1-r)(1-r) = 3$$(1-r)^2 = \frac{3}{27} = \frac{1}{9}$.Since it is a decreasing $G$.$P$.,$0 $r = 1 - \frac{1}{3} = \frac{2}{3}$.

The sum of the terms of an infinitely decreasing geometric progression is equal to the greatest value of the function $f(x) = x^3 + 3x - 9$ on the interval $[-2, 3]$. If the difference between the first and the second term of the progression is equal to $f'(0)$,then the common ratio of the $G.P.$ is

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