Question

Find the open interval where the function is increasing and decreasing. y=x^2+2x+3 (involves solving the derivative)

Answer 1

We have the next given function:

$y=x^2+2x+3$

The equation corresponds to a parabola.

First, we need to derivate it:

$\begin{gathered} \frac{d}{dx}y=\frac{d}{dx}(x^2+2x+3) \\ Then \\ \frac{d}{dx}y=\frac{d}{dx}x^2+\frac{d}{dx}2x+\frac{d}{dx}3 \\ \frac{d}{dx}y=2x+2+0 \\ \frac{d}{dx}y=2x+2 \end{gathered}$

Now, we need to find when x=0.

$\begin{gathered} 2x+2=0 \\ Solve\text{ for x} \\ 2x=-2 \\ x=-\frac{2}{2} \\ x=-1 \end{gathered}$

So, we can check the numbers using the number line:

Now, we need to check a number smaller than -1 and another number greater than -1.

Let us check 3.

Then:

$\begin{gathered} 2x+2 \\ 2(3)+2 \\ 6+2 \\ 8 \end{gathered}$

POSITIVE, when the x value is greater than -1, the function is positive (it is increasing).

Let us check -2:

$\begin{gathered} 2x+2 \\ 2(-2)+2 \\ -4+2 \\ -2 \end{gathered}$

NEGATIVE, when the x value is smaller than -1, the function is decreasing.

Finally, we can write each interval.

The function is decreasing on interval (-∞,-1)

The function is increasing on interval (-1,∞)