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Question: How do you estimate \(\vartriangle f\) using the linear approximation and use a calculator to comput...

How do you estimate △f\vartriangle f using the linear approximation and use a calculator to compute both the error and the percentage error given f(x)=18+xf(x) = \sqrt {18 + x} a=7a = 7 and △x=−0.5\vartriangle x = - 0.5?

Explanation

Solution

In this question we have been given with the error in the function and the function is to be calculated at f(7)f(7) therefore, we will use the formula of linear approximation by first finding out the derivative of the function and then find the percentage error from the value of the error.

Formula used: f(a+△x)=f(a)+f′(a)×△xf(a + \vartriangle x) = f(a) + f'(a) \times \vartriangle x

Complete step-by-step solution:
We have the given function as:f(x)=18+xf(x) = \sqrt {18 + x}
We have the error for the function as:△x=−0.5\vartriangle x = - 0.5
Now the formula for linear approximation is: f(a+△x)=f(a)+f′(a)×△xf(a + \vartriangle x) = f(a) + f'(a) \times \vartriangle x
Therefore, we will first find out f′(a)f'(a) by taking the derivative of the function.
we have: f(x)=18+xf(x) = \sqrt {18 + x}
The derivative of the term can be written as:
⇒f′(x)=d(18+x)dx\Rightarrow f'(x) = \dfrac{{d(\sqrt {18 + x} )}}{{dx}}
Now we know that ddxx=12x\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }} therefore, on using this formula, we get:
⇒f′(x)=1218+x\Rightarrow f'(x) = \dfrac{1}{{2\sqrt {18 + x} }}
Now we can write f(a)f(a) and f′(a)f'(a)as:
⇒f(a)=18+7\Rightarrow f(a) = \sqrt {18 + 7}
Which can be simplified as:
⇒f(a)=25\Rightarrow f(a) = \sqrt {25}
On taking the square root, we get:
⇒f(a)=5\Rightarrow f(a) = 5
Now f′(a)=1218+7f'(a) = \dfrac{1}{{2\sqrt {18 + 7} }}
On simplifying we get:
⇒f′(a)=1225\Rightarrow f'(a) = \dfrac{1}{{2\sqrt {25} }}
On taking the square root, we get:
⇒f′(a)=12×5\Rightarrow f'(a) = \dfrac{1}{{2 \times 5}}
On multiplying the terms in the denominator, we get:
⇒f′(a)=110\Rightarrow f'(a) = \dfrac{1}{{10}}
Now we have:
⇒f(7−0.5)=5+110×0.5\Rightarrow f(7 - 0.5) = 5 + \dfrac{1}{{10}} \times 0.5
On simplifying the expression by using a calculator, we get:
⇒f(7−0.5)=5.05\Rightarrow f(7 - 0.5) = 5.05
Now △f=f(7−0.5)−f(7)\vartriangle f = f(7 - 0.5) - f(7)
On substituting, we get:
⇒△f=5−5.05\Rightarrow \vartriangle f = 5 - 5.05
On subtracting, we get:
△f=−0.05\vartriangle f = - 0.05, which is the total error.
Now to find the percentage of error, we will use the formula of percentage which is: ⇒per=(partwhole×100)%\Rightarrow per = \left( {\dfrac{{part}}{{whole}} \times 100} \right)\%
Now on substituting the value of part=0.05part = 0.05 and whole=5whole = 5, which is the actual value of the function, we get:
⇒per=(0.055×100)%\Rightarrow per = \left( {\dfrac{{0.05}}{5} \times 100} \right)\%
On using the calculator, we get the error as:
⇒per=1%\Rightarrow per = 1\%
Therefore, the total percentage error is 1%1\% .

Note: It is to be remembered that the term error in mathematics is the difference between the true value and the approximate value of a term.
The most common types of errors in mathematics are round-off error and the truncation error.