Question

How do you find the intervals of increasing and decreasing using the first derivative given $y=x^2-2x-8$ ?

Answer 1

Hint:In the given question, we have to find the intervals in which a given function is increasing and decreasing by using the first derivative. The first derivative is defined as the differentiation of y with respect to x. A function is said to be increasing in a given interval if the value of y increases as we increase the value of x and the function is said to be decreasing if the value of y decreases on increasing the value of x. Using this information, we can find the correct answer.

Complete step by step answer:
We are given that $y=x^2-2x-8$
The first derivative of this function will be –
$$\begin{gathered} {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}(x^2-2x-8) \\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d(x^2)}{dx}}+{\displaystyle \frac{d(-2x)}{dx}}+{\displaystyle \frac{d(-8)}{dx}} \end{gathered}$$
We know that the differentiation of the product of a constant and a function is equal to the product of the constant and the derivative of the function, the derivative of $x^n$ is $n{x}^{n-1}$ and the the derivative of a constant is zero. So,
${\displaystyle \frac{dy}{dx}}=2x-2$
Now, in the increasing interval, the slope is positive, so –
$$\begin{gathered} {\displaystyle \frac{dy}{dx}}>0 \\ \Rightarrow 2x-2>0 \\ \Rightarrow 2x>2 \\ \Rightarrow x>1 \end{gathered}$$
And in the decreasing interval, the slope is negative, so –
$$\begin{gathered} {\displaystyle \frac{dy}{dx}}<0 \\ \Rightarrow 2x-2<0 \\ \Rightarrow 2x<2 \\ \Rightarrow x<1 \end{gathered}$$
Hence, the function $y=x^2-2x-8$ is increasing in the interval $(1,\mathrm{\infty })$ and decreasing in the interval $(-\mathrm{\infty },1)$ .

Note: The first derivative of a function represents its slope at any point. Thus, in the increasing interval, the function will have a curve going upwards, that is, the slope of the function in that interval will be positive, and in the decreasing interval the function will have a curve going downwards, that is, the slope of the function in that interval will be negative.