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

If $f(x) = kx - \sin x$ is monotonically increasing,then

Let $f(x) = \int \frac{x^2-3x+2}{x^4+1} \, dx$. Then the function decreases in the interval:

$f(x) = \cos x - 1 + \frac{x^2}{2!}, x \in R$. Then $f(x)$ is

Function $f(x) = |\sin x|$,$x \in \left(-\frac{\pi}{2}, 0\right)$ is . . . . . . .

Let $f : R \rightarrow R$ be defined as,
$f(x)=\begin{cases}-55 x, & \text{if } x<-5 \\ 2 x^{3}-3 x^{2}-120 x, & \text{if } -5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, & \text{if } x>4 \end{cases}$
Let $A=\{ x \in R : f \text{ is increasing} \}$. Then $A$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module