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

Feb 8, 2017

$\left(- \infty , 4\right)$.

#### Explanation:

Find the derivative using the chain rule. Let $y = \sqrt{u}$ and $u = 4 - x$. Then $\frac{\mathrm{du}}{\mathrm{dx}} = - 1$ and $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 \sqrt{u}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1 \cdot \frac{1}{2 \sqrt{u}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{4 - x}}$

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: $x = 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{4 - 3}} = - \frac{1}{2 \left(1\right)} = - \frac{1}{2}$

This means that in the interval $\left(- \infty , 4\right)$,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 $f \left(x\right)$ is decreasing on is $\left(- \infty , 4\right)$.

Hopefully this helps!