Respuesta :
Answer:
Approximation f(25.3)=5.03 (real value=5.0299)
The approximation can be written as f(x)=0.1x+2.5
Step-by-step explanation:
We have to approximate [tex]f(25.3)=\sqrt{25.3}[/tex] with a linear function.
To approximate a function, we can use the Taylor series.
[tex]f(x)=\sum_1^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n[/tex]
The point a should be a point where the value of f(a) is known or easy to calculate.
In this case, the appropiate value for a is a=25.
Then we calculate the Taylor series with a number of terms needed to make a linear estimation.
[tex]f(x)\approx f(a)+\frac{f'(a)}{1!}(x-a)[/tex]
The value of f'(a) needs the first derivate:
[tex]f'(x)=\frac{1}{2\sqrt{x}}\\\\f'(a)=f'(25)=\frac{1}{2\sqrt{25}}=\frac{1}{2*5}=\frac{1}{10}[/tex]
Then
[tex]f(x)\approx f(25)+\frac{1}{10}(x-25)=\sqrt{25} +\frac{1}{10}(x-25)\\\\f(x)\approx 5+\frac{1}{10}(x-25)[/tex]
We evaluate for x=25.3
[tex]f(25.3)\approx 5+\frac{1}{10}(25.3-25)\\\\f(25.3)\approx 5+\frac{1}{10}(0.3)=5.03[/tex]
If we rearrange the approximation to be in the form mx+b we have:
[tex]f(x)\approx 5+\frac{1}{10}(x-25)=5+0.1x-2.5\\\\f(x)\approx 0.1x+2.5[/tex]
Then, m=0.1 and b=2.5.
