Question

How do you find the largest open interval where the function is decreasing `f(x)=sqrt(4-x)`?

Answer 1

`(-oo, 4)`.

Explanation:

Find the derivative using the chain rule. Let `y = sqrt(u)` and `u = 4 - x`. Then `(du)/(dx)= -1` and `dy/(du)= 1/(2sqrt(u))`

`dy/dx = -1 * 1/(2sqrtu)`

`dy/dx= -1/(2sqrt(4 - x))`

The trick now is to find critical numbers. These will occur when the derivative equals `0` or is undefined. The derivative has a horizontal asymptote at `y = 0`, so there will be no point on the derivative where it equals `0`. It is undefined however at `x = 4`.

Now let's test to see which side is increasing and which side is decreasing, and accordingly, whether `x= 4` is an absolute maximum or an absolute minimum.

Test Point: `x = 3`

`dy/dx = -1/(2sqrt(4 - 3)) = -1/(2(1)) = -1/2`

This means that in the interval `(-oo, 4)`,the function is decreasing and that at `x= 4`, there is an absolute minimum. This is also an endpoint, as the function has domain `{x| x ≤ 4, x in RR}`.

Therefore, the largest open interval `f(x)` is decreasing on is `(-oo, 4)`.

Hopefully this helps!