Question

What is the average rate of change for the function for the interval, \[f\left( x \right)=\dfrac{3}{2-x}\] between x = 4 and x = 7?

Answer 1

Hint: 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 \[f\left( x \right)=\dfrac{3}{2-x}\] and we are asked to determine the average rate of change for this function between x = 4 and x =7.
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 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 = \[\displaystyle \lim_{\Delta x \to 0}\dfrac{\Delta f\left( x \right)}{\Delta x}\]
The above relation forms the basic concept of derivative of a function denoted as \[\dfrac{df\left( x \right)}{dx}\].
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\displaystyle \lim_{\Delta x \to 0}\dfrac{\Delta f\left( x \right)}{\Delta x}\]
Now, let us come to the question. Since we have to find the average rate of change, therefore we are going to use relation (1) to get our answer. Here, a = 4, b = 7, so we get,
\[\Rightarrow \Delta x=b-a=7-4=3\]
Substituting x = a = 4 in f (x) we get,
\[\Rightarrow f\left( a \right)=f\left( 4 \right)=\dfrac{-3}{4-2}=\dfrac{-3}{2}\]
Substituting x = b = 7 in f (x) we get,
\[\Rightarrow f\left( b \right)=f\left( 7 \right)=\dfrac{-3}{7-2}=\dfrac{-3}{5}\]
Therefore, change in the value of the function can be given as: -
\[\Rightarrow \Delta f\left( x \right)=f\left( b \right)-f\left( a \right)=\dfrac{-3}{5}+\dfrac{3}{2}=\dfrac{9}{10}\]
So, using relation (1) we get,
\[\Rightarrow \] Average rate of change of the function \[f\left( x \right)=\dfrac{\left( \dfrac{9}{10} \right)}{3}=\dfrac{9}{10\times 3}\]
$\therefore $ Average rate of change of the function \[f\left( x \right)=\dfrac{3}{10}\]

Note: 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 of x like in the above question. This is because instantaneous rate of change is found for infinitesimally small change in 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 obtained derivative.