Question

How do you find the average rate of change for the function $f\left( x \right)={{x}^{2}}-2x$ on the indicated intervals [1, 3]?

Answer 1

Assume ‘a’ and ‘b’ as the end – points of the given interval. Here ‘a’ is the lower endpoint and ‘b’ is the upper endpoint. Now, substitute the values of x as ‘a’ and ‘b’ one – by – one to find the values of the function at these endpoints, that is f (a) and f (b) respectively. Use the formula: - Average rate of change = $\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$ to get the answer.

Complete step by step answer:
Here, we have been provided with the function $f\left( x \right)={{x}^{2}}-2x$ and we are asked to determine the average rate of change for this function over the interval [1, 3].
Now, we know that there are two types of rate of change of a function, namely: - Instantaneous rate of change and average of change.
Average rate of a change of a function is the ratio of change of the value of the given function and the change in the value of the given function and the change in the value of the variable. Let us consider a function f (x) defined over the interval [a, b]. Here, ‘a’ and ‘b’ are respectively the lower and upper endpoint of the interval. Since, the variable is x therefore the average rate of change of f (x) is given as: -
$\Rightarrow$ Average rate of change = $\dfrac{\Delta f\left( x \right)}{\Delta x}$ - (1)
Here, $\Delta f\left( x \right)$ = change in f (x) = f (b) – f (a)
$\Rightarrow \Delta x$ = change in the variable = b – a
Now, Instantaneous rate of change of a function is defined as the rate of change in function when change in the variable tends to 0. Mathematically, it is denoted as: -
$\Rightarrow$ Instantaneous rate of change = $\underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{\Delta f\left( x \right)}{\Delta x}$
The above relation forms the basic concept of derivative of a function denoted as $\dfrac{df\left( x \right)}{dx}$.
$\Rightarrow \dfrac{df\left( x \right)}{dx}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{\Delta f\left( x \right)}{\Delta x}$
Now, let us come to the question. Since we have to find the average rate of change, therefore we are going to use the relation (1) to get our answer. We have,

& \Rightarrow f\left( x \right)={{x}^{2}}-2x \\\ & x\in \left[ 1,3 \right] \\\ \end{aligned}$$ Here, a = 1, b = 3 $$\Rightarrow \Delta x=b-a=3-1=2$$ Now, substituting x = a = 1 in f (x), we get, $$\Rightarrow f\left( a \right)=f\left( 1 \right)={{1}^{2}}-2\left( 1 \right)=-1$$ Substituting x = b = 3 in f (x), we get, $$\Rightarrow f\left( b \right)=f\left( 3 \right)={{3}^{2}}-2\left( 3 \right)=3$$ Therefore, change in the function can be given as: - $$\Rightarrow \Delta f\left( x \right)=f\left( b \right)-f\left( a \right)=3-\left( -1 \right)=4$$ So, using relation (1), we have, $$\Rightarrow $$ Average rate of change of $$f\left( x \right)=\dfrac{4}{2}$$ $$\Rightarrow $$ Average rate of change of $$f\left( x \right)=2$$ Hence, the average rate of change for the function $$f\left( x \right)={{x}^{2}}-2x$$ on the interval [1, 3] is 2. **Note:** One may note that if we would have been asked to determine the instantaneous rate of change of f (x) then we would not have been provided with an interval like in the above question. This is because instantaneous rate of change is found for infinitesimally small change for the value of x, i.e., $$\Delta x\to 0$$. In such a case, to find the answer we need to differentiate the function f (x) and find $$\dfrac{df\left( x \right)}{dx}$$ and then finally we need to substitute the value of x in the expression of $$\dfrac{df\left( x \right)}{dx}$$. You must remember these both types of rate of change because these topics mark the starting of the topic ‘calculus’ which is used everywhere in the field of mathematics.