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Question: If \(f(x) = {x^2}\) then find the value of \(\dfrac{{f(1.1) - f(1)}}{{1.1 - 1}}\)....

If f(x)=x2f(x) = {x^2} then find the value of f(1.1)−f(1)1.1−1\dfrac{{f(1.1) - f(1)}}{{1.1 - 1}}.

Explanation

Solution

We have given the function f(x) then using this function we will find the values that will be required in finding the solution i.e. finding the value of the expression f(1.1)−f(1)1.1−1\dfrac{{f(1.1) - f(1)}}{{1.1 - 1}}, so basically we will find f(1.1)f(1.1) and f(1)f(1), then substituting those values in the given expression we will get the answer that is required.

Complete step by step solution: Given f(x)=x2f(x) = {x^2}

f(x)=x2f(x) = {x^2}
Now, Substituting x=1.1x = 1.1, we get,
⇒f(1.1)=(1.1)2\Rightarrow f(1.1) = {(1.1)^2}
⇒f(1.1)=(1.1)(1.1)\Rightarrow f(1.1) = (1.1)(1.1)
On squaring we get,
∴f(1.1)=1.21\therefore f(1.1) = 1.21
Now, substituting x=1x = 1 in f(x)=x2f(x) = {x^2}, we get,
⇒f(1)=(1)2\Rightarrow f(1) = {(1)^2}
On squaring we get,
∴f(1)=1\therefore f(1) = 1
Now substituting the value of f(1.1)f(1.1) and f(1)f(1) in the expression which we have to find i.e. f(1.1)−f(1)1.1−1\dfrac{{f(1.1) - f(1)}}{{1.1 - 1}}
⇒f(1.1)−f(1)1.1−1=1.21−11.1−1\Rightarrow \dfrac{{f(1.1) - f(1)}}{{1.1 - 1}} = \dfrac{{1.21 - 1}}{{1.1 - 1}}
On Simplifying the numerator and the denominator, we get,
⇒f(1.1)−f(1)1.1−1=0.210.1\Rightarrow \dfrac{{f(1.1) - f(1)}}{{1.1 - 1}} = \dfrac{{0.21}}{{0.1}}
On Multiplying 10 on numerator and denominator, we get,
⇒f(1.1)−f(1)1.1−1=2.11\Rightarrow \dfrac{{f(1.1) - f(1)}}{{1.1 - 1}} = \dfrac{{2.1}}{1}

Therefore, f(1.1)−f(1)1.1−1=2.1\dfrac{{f(1.1) - f(1)}}{{1.1 - 1}} = 2.1

Note: Note: An alternative method for the solution of this question can be using the derivative as we know that f(x)−f(y)x−y\dfrac{{f(x) - f(y)}}{{x - y}} is the slope of the graph between f(x) and f(y) but the function we have given in the equation of do not have a constant slope and changes with every value of x so we’ll write it as the average of the slopes at x and y i.e. f(x)−f(y)x−y=f′(x)+f′(y)2\dfrac{{f(x) - f(y)}}{{x - y}} = \dfrac{{f'(x) + f'(y)}}{2}.

Therefore we can say the value of f(1.1)−f(1)1.1−1\dfrac{{f(1.1) - f(1)}}{{1.1 - 1}} will b equal to f′(1.1)+f′(1)2\dfrac{{f'(1.1) + f'(1)}}{2}
On differentiating f(x)
f′(x)=2xf'(x) = 2x
Substituting x=1.1
⇒f′(1.1)=2(1.1)\Rightarrow f'(1.1) = 2(1.1)
∴f′(1.1)=2.2\therefore f'(1.1) = 2.2
Substituting x=1
⇒f′(1)=2(1)\Rightarrow f'(1) = 2(1)
∴f′(1)=2\therefore f'(1) = 2
Now, f′(1.1)+f′(1)2=2.2+22\dfrac{{f'(1.1) + f'(1)}}{2} = \dfrac{{2.2 + 2}}{2}
=4.22= \dfrac{{4.2}}{2}
=2.1= 2.1, which is giving the same result as the above solution.