# How do you find the intervals of increasing and decreasing using the first derivative given y=(x-1)^2(x+3)?

Oct 1, 2016

The function $y = {\left(x - 1\right)}^{2} \left(x + 3\right)$ is rising in the interval $\left(- \infty , - \frac{5}{3}\right)$ and then it starts declining in the interval $\left(- \frac{5}{3} , 1\right)$ and then again rising in the interval $\left(1 , \infty\right)$

#### Explanation:

As $y = {\left(x - 1\right)}^{2} \left(x + 3\right)$

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

= $2 \left({x}^{2} + 3 x - x - 3\right) + {x}^{2} - 2 x + 1$

= $3 {x}^{2} + 2 x - 5$

= $3 {x}^{2} - 3 x + 5 x - 5$

= $3 x \left(x - 1\right) + 5 \left(x - 1\right)$

= $\left(3 x + 5\right) \left(x - 1\right)$ whose zeros are $- \frac{5}{3}$ and $1$

Now using Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - \frac{5}{3} \textcolor{w h i t e}{X X X X X} 1$

$\left(x - 1\right) \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left(3 x + 5\right) \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left(3 x + 5\right) \left(x - 1\right) \textcolor{w h i t e}{} + i v e \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

Hence, the function $y = {\left(x - 1\right)}^{2} \left(x + 3\right)$ is rising in the interval $\left(- \infty , - \frac{5}{3}\right)$ and then it starts declining in the interval $\left(- \frac{5}{3} , 1\right)$ and then again rising in the interval $\left(1 , \infty\right)$

$- \frac{5}{3}$ amd $1$ are local maxima and minima.

graph{(x-1)^2(x+3) [-5, 5, -10, 10]}