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

In the interval $(0, 1)$,the maximum value of the function $f(x) = |x \ln x|$ is:

Let $f(x) = \int_0^{x^2} \frac{t^2-8t+15}{e^t} dt$,$x \in R$. Then the numbers of local maximum and local minimum points of $f$,respectively,are:

The curve $y(x) = ax^{3} + bx^{2} + cx + 5$ touches the $x$-axis at the point $P(-2, 0)$ and cuts the $y$-axis at the point $Q$,where the derivative $y'(0) = 3$. Find the local maximum value of $y(x)$.

Find the local maximum and local minimum values for the function $g(x) = \frac{1}{x^{2}+2}$.

If $m$ and $M$ are the absolute minimum and absolute maximum values of the function $f(x) = 2\sqrt{2} \sin x - \tan x$ in the interval $[0, \pi/3]$,then $m + M =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module