The function f(x) = x^4 - frac{x^3}{3} is: | vedclass

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

Question

(A) Given the function $f(x) = x^4 - \frac{x^3}{3}$.To determine the intervals of increase and decrease,we find the first derivative $f'(x)$:$f'(x) =

Vedclass · Accepted Answer

Increasing for $x > \frac{1}{4}$ and decreasing for $x < \frac{1}{4}$

Vedclass · Answer

(A) Given the function $f(x) = x^4 - \frac{x^3}{3}$.To determine the intervals of increase and decrease,we find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^4 - \frac{x^3}{3}) = 4x^3 - x^2$.Set $f'(x) = 0$ to find the critical points:$x^2(4x - 1) = 0$,which gives $x = 0$ and $x = \frac{1}{4}$.For the function to be increasing,$f'(x) > 0$:$x^2(4x - 1) > 0$.Since $x^2$ is always non-negative,$f'(x) > 0$ when $4x - 1 > 0$,which means $x > \frac{1}{4}$.For the function to be decreasing,$f'(x) $x^2(4x - 1) Thus,the function is increasing for $x > \frac{1}{4}$ and decreasing for $x < \frac{1}{4}$.

The function $f(x) = x^4 - \frac{x^3}{3}$ is:

Similar Questions

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

Let $f(x) = \int\limits_1^x {\left( {t\ln(t) - \frac{{\ln(t)}}{t}} \right)dt}$ for $x > 1$. Then:

The set of all real values of $a$ such that the real valued function $f(x) = x^3 + 2ax^2 + 3(a+1)x + 5$ is strictly increasing in its entire domain is

The derivative of the function $f(x) = \cos^4 x + \sin^4 x$ for $0 \le x \le 2\pi$ is positive for:

Let $f(x) = xe^{x(1 - x)}$. Then $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module