Question

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

Answer 1

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 –

$$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}({x^2} - 2x - 8) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d({x^2})}}{{dx}} + \dfrac{{d( - 2x)}}{{dx}} + \dfrac{{d( - 8)}}{{dx}} \\\$$

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,
$\dfrac{{dy}}{{dx}} = 2x - 2$
Now, in the increasing interval, the slope is positive, so –
$\dfrac{{dy}}{{dx}} > 0 \\\ \Rightarrow 2x - 2 > 0 \\\ \Rightarrow 2x > 2 \\\ \Rightarrow x > 1 \\\$
And in the decreasing interval, the slope is negative, so –
$\dfrac{{dy}}{{dx}} < 0 \\\ \Rightarrow 2x - 2 < 0 \\\ \Rightarrow 2x < 2 \\\ \Rightarrow x < 1 \\\$
Hence, the function $y = {x^2} - 2x - 8$ is increasing in the interval $(1,\infty )$ and decreasing in the interval $( - \infty ,1)$ .

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.