If f(x) = x^{3/2}(3x - 10),x geq 0,then in wh | vedclass

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

Question

(B) Given function is $f(x) = x^{3/2}(3x - 10) = 3x^{5/2} - 10x^{3/2}$.To find the interval where the function is increasing,we find the derivative $f

Vedclass · Accepted Answer

$[2, \infty)$

Vedclass · Answer

(B) Given function is $f(x) = x^{3/2}(3x - 10) = 3x^{5/2} - 10x^{3/2}$.To find the interval where the function is increasing,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(3x^{5/2} - 10x^{3/2}) = 3 \cdot \frac{5}{2} x^{3/2} - 10 \cdot \frac{3}{2} x^{1/2} = \frac{15}{2} x^{3/2} - 15 x^{1/2}$.Setting $f'(x) = 0$ for critical points:$\frac{15}{2} x^{1/2} (x - 2) = 0$.Since $x \geq 0$,the critical point is $x = 2$.For $x > 2$,$f'(x) > 0$,so the function is increasing on $[2, \infty)$.For $0 < x < 2$,$f'(x) < 0$,so the function is decreasing on $(0, 2)$.

If $f(x) = x^{3/2}(3x - 10)$,$x \geq 0$,then in which interval is $f(x)$ an increasing function?

Similar Questions

If $f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x)$,then $f(x)$

For what values of $x$ is the function $f(x) = [x(x - 3)]^2$ an increasing function?

In the interval $(7, \infty)$,the function $f(x) = |x-5| + 2|x-7|$ is:

Let $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19$. Then, $f(x)=0$ has

If the function $y = \sin x(1 + \cos x)$ is defined in the interval $[-\pi, \pi]$,then $y$ is strictly increasing in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module