Find the minimum of this function

Calculus Level 2

Find the minimum value of the function f ( x ) = x + 1 x f(x) = x + \dfrac{1}{x} , given that x > 0 x > 0 .

1 2 0 -\infty

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

. .
Apr 2, 2021

Substituting x = 1 x = 1 , then f ( x ) = 1 + 1 = 2 f ( x ) = 1 + 1 = 2 .

By AM-GM Inequality, \text{By AM-GM Inequality,} x + 1 x 2 x × 1 x x+\dfrac{1}{x} \geq 2\sqrt{x\times \dfrac{1}{x} } x > 0 x × 1 x = 1 x>0 \implies \sqrt{x\times \dfrac{1}{x} } =1 min ( x + 1 x ) = 2 \implies \min\left(x+\dfrac{1}{x}\right) =2

Srinivasa Gopal
Oct 7, 2017

f(x) = ( x + (1/x)); d f ( x ) d x \frac{df(x)}{dx} = 1 - 1 x x \frac{1}{x*x} ; Min occurs when f '(x) = 0 and when f ''(x) is positive. Calculating f '(x) ( 1 -( 1/x*x)) = 0 occurs at either x = 1 or x = -1. f ''(x) = 2/x^3. f ''(1) = 2/3; f ''(-1) = (-2/3). Hence we can conclude that minimum occurs at x = 1 , follows that f(1) = 2;
The minimum value of f(x) = 2.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...