Let f(x)=x^2+frac{1}{x^2} and g(x)=x-frac{1}{ | vedclass

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

Question

(B) Given $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$.We can write $f(x)$ in terms of $g(x)$ as $f(x)=\left(x-\frac{1}{x}\right)^2+2$.Let $t=x-\

Vedclass · Accepted Answer

$2 \sqrt{2}$

Vedclass · Answer

(B) Given $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$.We can write $f(x)$ in terms of $g(x)$ as $f(x)=\left(x-\frac{1}{x}\right)^2+2$.Let $t=x-\frac{1}{x}$. Then $f(x)=t^2+2$ and $g(x)=t$.We define $h(t)=\frac{f(x)}{g(x)}=\frac{t^2+2}{t}=t+\frac{2}{t}$.To find the local minimum,we find the derivative $h'(t)=1-\frac{2}{t^2}$.Setting $h'(t)=0$,we get $1-\frac{2}{t^2}=0$,which implies $t^2=2$,so $t=\pm \sqrt{2}$.Using the second derivative test,$h''(t)=\frac{4}{t^3}$.For $t=\sqrt{2}$,$h''(\sqrt{2})=\frac{4}{(\sqrt{2})^3} > 0$,so $t=\sqrt{2}$ is a point of local minimum.The local minimum value is $h(\sqrt{2})=\sqrt{2}+\frac{2}{\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$.

Let $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$ for $x \in R-\{-1,0,1\}$,then the local minimum of $\frac{f(x)}{g(x)}$ is

Similar Questions

Let the radius and height of a right circular cylinder be related as $r^2 + h = 6$. If the volume of the cylinder is maximum,then the value of $\frac{r}{h}$ is:

Prove that the function $f(x) = e^{x}$ does not have any maxima or minima.

The maximum value of the function $f(x)=3x^3-18x^2+27x-40$ on the set $S=\{x \in R : x^2+30 \leq 11x\}$ is

If the lengths of three sides of a trapezium other than the base are equal to $10 \ cm$,find the maximum area of the trapezium.

If the local maximum value of the function $f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}, \quad x \in\left(0, \frac{\pi}{2}\right)$,is $\frac{k}{e}$,then $\left(\frac{ k }{ e }\right)^8+\frac{ k ^8}{ e ^5}+ k ^8$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module