Find the open intervals on x-axis on which the function f(x)= x^2/x-1 is increasing and those on which it is?

Question

decreasing

Answer 1

3

Answer 2

(negative infinity,0) increasing
(0,1) decreasing
1 undefined
(1,2) decreasing
(2, positive infinity)

The first derivative of (x^2)/(x-1) = (x^2-2x)/(x-1)^2
or [x(x-2)]/(x-1)^2
The first derivative is used to find the slope of the tangent line, so I found that the tangent is 0 at x=0 and x=2. I also found that for values less than 0 and greater than 2 the tangent has a positive slope. I also found that between 0 and 2 the slope of the tangent line is negative. The function and all its derivatives are undefined at x=1 because division by 0 in undefined.

Answer 3

Begin by defining your limits, ie. x cannot equal 1 since that would make your denominator 0.

Now put a number smaller than your limit (1) into the equation and see what happens to your f(x), ie if x=0.1, then f(x)=0.01/-0.9 (I don't have my calc. handy) which is a small negative number.

This means that for numbers <1, the function is decreasing

Now put a number larger than 1 into the equation and see what happens to your f(x), ie if x=2, then f(x)=4

This means for all numbers >1, f(x) is increasing

-infinity x>+infinity increasing, vertical asymptote at x=1

Hope this makes sense!