The function,f(x)=(3x-7)x^{2/3}, x in R, is i | vedclass

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

Question

(B) Given function: $f(x) = (3x - 7)x^{2/3} = 3x^{5/3} - 7x^{2/3}$.To find the intervals where the function is increasing,we find the derivative $f'(x

Vedclass · Accepted Answer

$(-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)$

Vedclass · Answer

(B) Given function: $f(x) = (3x - 7)x^{2/3} = 3x^{5/3} - 7x^{2/3}$.To find the intervals where the function is increasing,we find the derivative $f'(x)$:$f'(x) = 3 \cdot \frac{5}{3}x^{2/3} - 7 \cdot \frac{2}{3}x^{-1/3}$$f'(x) = 5x^{2/3} - \frac{14}{3x^{1/3}}$Simplify the expression for $f'(x)$:$f'(x) = \frac{5x^{2/3} \cdot 3x^{1/3} - 14}{3x^{1/3}} = \frac{15x - 14}{3x^{1/3}}$For the function to be increasing,we require $f'(x) > 0$:$\frac{15x - 14}{3x^{1/3}} > 0$We analyze the sign of $f'(x)$ using the critical points $x = 0$ and $x = \frac{14}{15}$:- For $x  0$.- For $0  0$,so $f'(x) - For $x > \frac{14}{15}$,$15x - 14 > 0$ and $3x^{1/3} > 0$,so $f'(x) > 0$.Thus,$f(x)$ is increasing for $x \in (-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)$.

The function,$f(x)=(3x-7)x^{2/3}, x \in R,$ is increasing for all $x$ lying in

Similar Questions

If $y = 8x^3 - 60x^2 + 144x + 27$ is a decreasing function on the interval $(a, b)$,then $(a, b) = $

If $f''(x) < 0$ for all $x \in (0, 2)$,then the function $H(x) = f(1 - x) + 2f(x/2)$ is:

If $f$ is a real-valued differentiable function such that $f(x) f^{\prime}(x) < 0$ for all real $x,$ then

Show that the function $f$ given by $f(x) = \tan^{-1}(\sin x + \cos x), x > 0$ is always an increasing function in $\left(0, \frac{\pi}{4}\right)$.

For every value of $x$,the function $f(x) = \frac{1}{5^x}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module