Let y= x + 1/x ,x not =0. Find the intervals on which f is an increasing function?

Question

Let y= x + 1/x ,x not =0. Find the intervals on which f is an increasing function?

2 Answers

Impact of this question

1125 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-09T14:05:48

$y = x + \frac{1}{x}$ $y = x+1/x$ is increasing in the intervals $(- \infty, - 1) \cup (1, + \infty)$ $(-oo,-1) uu (1,+oo)$

Explanation:

We're looking for the intervals where $\frac{d y}{d x} > 0$ $dy/dx > 0$ .
Recall that a function is increasing at a point when it's derivative at that point is greater than zero.

Getting the derivative
First, we get the derivative of $x + \frac{1}{x}$ $x+1/x$ .

$\frac{d}{d x} (x + \frac{1}{x})$ $d/dx (x+1/x)$

$= \frac{d}{d x} x + \frac{d}{d x} (x^{- 1})$ $=d/dxx + d/dx(x^-1)$

$= (1) + (- 1) (x^{- 2})$ $=(1) + (-1)(x^-2)$

$= 1 - \frac{1}{x^{2}}$ $=1-1/(x^2)$

$\frac{d y}{d x} = 1 - \frac{1}{x^{2}}$ $dy/dx = 1-1/x^2$

Now, we need to solve for:

$1 - \frac{1}{x^{2}} > 0$ $1-1/(x^2) > 0$

$1 > \frac{1}{x^{2}}$ $1> 1/(x^2)$

Solving for the intervals (one approach)

Since $x^{2}$ $x^2$ is always positive for any real number $x$ $x$ , we can multiply both sides by $x^{2}$ $x^2$ without changing the sign.

$x^{2} > 1$ $x^2> 1$

$x^{2} - 1 > 0$ $x^2 - 1> 0$

The zeroes of $x^{2} - 1$ $x^2-1$ are ${1, - 1}$ ${1,-1}$ , and the parabola curves upward, since the lead coefficient is greater than $0$ $0$ .

This results in the graph:

graph{x^2-1 [-2.624, 2.85, -1.727, 1.01]}

Therefore $\frac{d y}{d x} > 0$ $dy/dx > 0$ over the intervals $(- \infty, - 1) \cup (1, + \infty)$ $(-oo,-1) uu (1,+oo)$ .

So, the equation is increasing over those intervals.

Answer 2 · 2015-09-10T08:57:45

The answer can be put into a compact form

Explanation:

Can be put into a compact form as below:

enter image source here

Science

Math

Humanities

... and beyond

Let y= x + 1/x ,x not =0. Find the intervals on which f is an increasing function?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question