For all x in mathbb{R}, the minimum value fra | vedclass

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

Question

(A) Let $y = \frac{x^2+x+1}{x^2-x+1}$.To find the critical points, we differentiate $y$ with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $y = \frac{x^2+x+1}{x^2-x+1}$.To find the critical points, we differentiate $y$ with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(x^2-x+1)(2x+1) - (x^2+x+1)(2x-1)}{(x^2-x+1)^2}$$= \frac{(2x^3 - 2x^2 + 2x + x^2 - x + 1) - (2x^3 + 2x^2 + 2x - x^2 - x - 1)}{(x^2-x+1)^2}$$= \frac{(2x^3 - x^2 + x + 1) - (2x^3 + x^2 + x - 1)}{(x^2-x+1)^2}$$= \frac{-2x^2 + 2}{(x^2-x+1)^2} = \frac{2(1-x^2)}{(x^2-x+1)^2}$.Setting $\frac{dy}{dx} = 0$, we get $1-x^2 = 0$, which implies $x = 1$ or $x = -1$.For $x = 1$, $y = \frac{1+1+1}{1-1+1} = 3$ (Maximum value).For $x = -1$, $y = \frac{1-1+1}{1+1+1} = \frac{1}{3}$ (Minimum value).Thus, the minimum value $\frac{1}{3}$ occurs at $l = -1$ and the maximum value $3$ occurs at $m = 1$.Therefore, $l+m = -1 + 1 = 0$.

For all $x \in \mathbb{R}$, the minimum value $\frac{1}{3}$ and the maximum value $3$ of $f(x) = \frac{x^2+x+1}{x^2-x+1}$ occur at $l$ and $m$ respectively. Then $l+m$ is equal to:

Similar Questions

The minimum value of the function $f(x) = x \log x$ is

Let $f(x) = \begin{cases} |x|, & 0 < |x| \le 2 \\ 1, & x = 0 \end{cases}$,then at $x = 0$,$f$ has

Maximum value of the function $f(x)=\frac{x}{8}+\frac{2}{x}$ on the interval $[1,6]$ is

Find both the maximum value and the minimum value of $3x^{4}-8x^{3}+12x^{2}-48x+25$ on the interval $[0,3]$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module