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

(B) Let $y = \frac{1-x+x^{2}}{1+x+x^{2}}$.We can rewrite the expression as $y = \frac{(x^{2}+x+1) - 2x}{x^{2}+x+1} = 1 - \frac{2x}{x^{2}+x+1}$.To find

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(B) Let $y = \frac{1-x+x^{2}}{1+x+x^{2}}$.We can rewrite the expression as $y = \frac{(x^{2}+x+1) - 2x}{x^{2}+x+1} = 1 - \frac{2x}{x^{2}+x+1}$.To find the minimum value of $y$,we need to maximize the term $\frac{2x}{x^{2}+x+1}$.Let $f(x) = \frac{x}{x^{2}+x+1}$.Taking the derivative with respect to $x$ and setting it to $0$:$f'(x) = \frac{(x^{2}+x+1)(1) - x(2x+1)}{(x^{2}+x+1)^{2}} = \frac{x^{2}+x+1-2x^{2}-x}{(x^{2}+x+1)^{2}} = \frac{1-x^{2}}{(x^{2}+x+1)^{2}}$.Setting $f'(x) = 0$ gives $1-x^{2} = 0$,so $x = 1$ or $x = -1$.For $x = 1$,$f(1) = \frac{1}{1+1+1} = \frac{1}{3}$.For $x = -1$,$f(-1) = \frac{-1}{1-1+1} = -1$.Since we want to minimize $y = 1 - 2f(x)$,we choose the maximum value of $f(x)$,which is $1/3$.Thus,the minimum value of $y$ is $1 - 2(1/3) = 1 - 2/3 = 1/3$.

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

Similar Questions

For $x + y = 8$,what is the maximum value of $xy$?

On the interval $[1, e]$,the greatest value of $f(x) = x^2 \log x$ is:

Find the absolute maximum of $x^{40}-x^{20}$ on the interval $[0,1]$.

Number of solutions of the equation $3 \tan x + x^3 = 2$ in the interval $(0, \frac{\pi}{4})$ is:

The number $10$ is divided into two parts such that the sum of double of the first part and the square of the second part is minimum. The two parts are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module