If f(x) = x + frac{1}{x},x neq 0,then the loc | vedclass

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

Question

(B) Given $f(x) = x + \frac{1}{x}$,$x 
eq 0$. Differentiating with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2}$. For local maxima or minima,we s

Vedclass · Accepted Answer

$-2$ and $2$

Vedclass · Answer

(B) Given $f(x) = x + \frac{1}{x}$,$x 
eq 0$. Differentiating with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2}$. For local maxima or minima,we set $f'(x) = 0$. $1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1, -1$. Now,we check the sign of $f'(x)$ around these points: For $x  0$. For $-1 The local maximum value is $f(-1) = -1 + \frac{1}{-1} = -2$. For $0  1$,$f'(x) > 0$. Since $f'(x)$ changes from negative to positive at $x = 1$,$x = 1$ is a point of local minima. The local minimum value is $f(1) = 1 + \frac{1}{1} = 2$. Thus,the local maximum value is $-2$ and the local minimum value is $2$.

If $f(x) = x + \frac{1}{x}$,$x \neq 0$,then the local maximum and minimum values of the function $f$ are respectively....

Similar Questions

If $f(x) = 7e^{\sin^2 x} - e^{\cos^2 x} + 2$,then $\sqrt{7f_{\min} + f_{\max}}$ is equal to

Find the absolute maximum and minimum values of the function $f$ given by $f(x) = \cos^{2} x + \sin x$,where $x \in [0, \pi]$.

On the interval $[0,1]$,the function $f(x) = x^{25}(1-x)^{75}$ takes its maximum value at the point

At which point in the interval $[0, 1]$ is the function $f(x) = x^{25}(1 - x)^{75}$ maximum?

If $x - 2y = 4,$ the minimum value of $xy$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module