For all real x,the minimum value of frac{1-x+ | vedclass

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

Question

(C) Let $f(x) = \frac{1-x+x^2}{1+x+x^2}$.Using the quotient rule,$f'(x) = \frac{(1+x+x^2)(-1+2x) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}$.Expanding the numera

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) Let $f(x) = \frac{1-x+x^2}{1+x+x^2}$.Using the quotient rule,$f'(x) = \frac{(1+x+x^2)(-1+2x) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}$.Expanding the numerator: $(1+x+x^2)(-1+2x) = -1+2x-x+2x^2-x^2+2x^3 = 2x^3+x^2+x-1$.$(1-x+x^2)(1+2x) = 1+2x-x-2x^2+x^2+2x^3 = 2x^3-x^2+x+1$.Subtracting these: $(2x^3+x^2+x-1) - (2x^3-x^2+x+1) = 2x^2-2$.Thus,$f'(x) = \frac{2x^2-2}{(1+x+x^2)^2}$.Setting $f'(x) = 0$ gives $2x^2-2 = 0$,so $x^2 = 1$,which means $x = 1$ or $x = -1$.Evaluating $f(x)$ at these points:$f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$.$f(-1) = \frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = \frac{3}{1} = 3$.Therefore,the minimum value of $f(x)$ is $\frac{1}{3}$.

For all real $x$,the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

Similar Questions

The number of critical points of the function $f(x)=(x-2)^{2/3}(2x+1)$ is:

Show that of all the rectangles inscribed in a given fixed circle,the square has the maximum area.

Let $f(x)=3^{(x^{2}-2)^{3}+4}, x \in R$. Then which of the following statements are true?
$P: x=0$ is a point of local minima of $f$
$Q: x=\sqrt{2}$ is a point of inflection of $f$
$R: f^{\prime}$ is increasing for $x>\sqrt{2}$

The height of a right circular cone of maximum volume inscribed in a sphere of diameter $a$ is

$A$ box is to be made with a square base and an open top. If the area of the material used is $48 \, m^2$,then the maximum volume of the box is ........... $m^3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module