# How do you find the intervals of increasing and decreasing using the first derivative given y=(x-1)^(1/3)?

Nov 21, 2016

Differentiate.

Let $y = {u}^{\frac{1}{3}}$ and $u = x - 1$.

$y ' = \frac{1}{3} {u}^{- \frac{2}{3}}$ and $u ' = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 {u}^{\frac{2}{3}}}$

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

The derivative will have a critical value of $1$, since it renders the derivative undefined. Check to the left and right of this point.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 {\left(3 - 1\right)}^{\frac{2}{3}}} = \frac{1}{3 \left(\sqrt[3]{4}\right)} \to \text{ positive: the function is " color(blue) "increasing}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 {\left(- 2 - 1\right)}^{\frac{2}{3}}} = \frac{1}{3 \left(\sqrt[3]{9}\right)} \to \text{ positive: the function is "color(blue) "increasing}$

So, the function $y = {\left(x - 1\right)}^{\frac{1}{3}}$ is uniformly increasing.

Hopefully this helps!