# How do you find the intervals of increasing and decreasing using the first derivative given y=-2x^2+4x+3?

Jun 16, 2017

$f \left(x\right)$ is increasing from $\left(- \infty , 1\right)$
$f \left(x\right)$ is decreasing from $\left(1 , \infty\right)$

#### Explanation:

We want to perform that first derivative test here:
We begin by differentiate using the power rule:

$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

$\frac{d}{\mathrm{dx}} = - 2 \left(2\right) {x}^{2 - 1} + 4 \left(1\right) {x}^{\cancel{1 - 1}} + 0$

Keep in mind that ${x}^{0} = 1$ and that derivative of a constant is zero.

$f ' \left(x\right) = - 4 x + 4$

Now we want to factor and set it equal to zero:

$- 4 \left(x - 1\right) = 0$

$x - 1 = 0$

$x = 1$

We create a test a interval from $\left(- \infty , 1\right) \cup \left(1 , \infty\right)$
Now you pick numbers in between the interval and test them in the derivative. If the number is positive this means the function is increasing and if it's negative the function is decreasing.

I picked 0 a number from the left

$f ' \left(0\right) = 4$

This means from $\left(\infty , 1\right)$ the function is increasing.

Then I picked a number from the right which was 2.

$f ' \left(2\right) = - 4$

This means from $\left(- 1 , \infty\right)$ the function is decreasing.

So, from $\left(\infty , 1\right)$ the function is increasing and from $\left(- 1 , \infty\right)$ the function is decreasing.

Note: For this exact reason we can say that there's an absolute max at $f \left(1\right)$. We can say this because its only a parabola.