The number of critical points of the function | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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:

Similar Questions

The maximum value of the function $f(x) = -|x+1| + 3, x \in R$ is . . . . . . .

$A$ triangular garden is fenced on two sides,and the third side is a straight river bank. The two fenced sides have an equal length of $x$. What is the maximum area that can be enclosed by the garden?

If a right circular cone having maximum volume is inscribed in a sphere of radius $3 \, cm$,then the curved surface area (in $cm^2$) of this cone is

If $a^2 x^4 + b^2 y^4 = c^6$,then the maximum value of $xy$ is:

$A$ wire of length $20 \ m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in $meters$) of the hexagon,so that the combined area of the square and the hexagon is minimum,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module