The maximum value of $f(x) = \frac{x}{4 + x + x^2}$ on the interval $[-1, 1]$ is:

  • A
    $-\frac{1}{4}$
  • B
    $-\frac{1}{3}$
  • C
    $\frac{1}{6}$
  • D
    $\frac{1}{5}$

Explore More

Similar Questions

The value of the function $f(x) = (x - 1)(x - 2)^2$ at its maxima is

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$,then

What are the minimum and maximum values of the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$?

The minimum value of the function $f(x) = 2 x^3 - 15 x^2 + 36 x - 48$ on the set $A = \{x \mid x^2 + 20 \le 9 x\}$ is

For the function $f(x) = x \cos \frac{1}{x}, \quad x \geq 1$,consider the following statements:
$(A)$ For at least one $x$ in the interval $[1, \infty), f(x+2)-f(x) < 2$
$(B)$ $\lim _{x \rightarrow \infty} f^{\prime}(x) = 1$
$(C)$ For all $x$ in the interval $[1, \infty), f(x+2)-f(x) > 2$
$(D)$ $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$
Which of the following combinations of statements is correct?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo