The number of critical points of the function | vedclass

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(A) Given function: $f(x) = (x-2)^{2/3}(2x+1)$.To find the critical points,we calculate the derivative $f'(x)$ using the product rule:$f'(x) = \frac{d

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$2$

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(A) Given function: $f(x) = (x-2)^{2/3}(2x+1)$.To find the critical points,we calculate the derivative $f'(x)$ using the product rule:$f'(x) = \frac{d}{dx}[(x-2)^{2/3}] \cdot (2x+1) + (x-2)^{2/3} \cdot \frac{d}{dx}[2x+1]$$f'(x) = \frac{2}{3}(x-2)^{-1/3}(2x+1) + 2(x-2)^{2/3}$Factor out $2(x-2)^{-1/3}$:$f'(x) = 2(x-2)^{-1/3} [\frac{1}{3}(2x+1) + (x-2)]$$f'(x) = \frac{2}{3(x-2)^{1/3}} [2x + 1 + 3x - 6]$$f'(x) = \frac{2(5x-5)}{3(x-2)^{1/3}} = \frac{10(x-1)}{3(x-2)^{1/3}}$Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined.$f'(x) = 0 \implies x-1 = 0 \implies x = 1$.$f'(x)$ is undefined when the denominator is zero,i.e.,$x-2 = 0 \implies x = 2$.Thus,the critical points are $x=1$ and $x=2$.There are $2$ critical points.

The number of critical points of the function $f(x)=(x-2)^{2/3}(2x+1)$ is:

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