Question

Use linear approximation to approximate 25.3‾‾‾‾√ as follows. Let f(x)=x√. The equation of the tangent line to f(x) at x=25 can be written in the form y=mx+b. Compute m and b.

Answer 1

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 $f(25.3)=\sqrt{25.3}$ with a linear function.

To approximate a function, we can use the Taylor series.

$f(x)=\sum_1^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$

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.

$f(x)\approx f(a)+\frac{f'(a)}{1!}(x-a)$

The value of f'(a) needs the first derivate:

$f'(x)=\frac{1}{2\sqrt{x}}\\\\f'(a)=f'(25)=\frac{1}{2\sqrt{25}}=\frac{1}{2*5}=\frac{1}{10}$

Then

$f(x)\approx f(25)+\frac{1}{10}(x-25)=\sqrt{25} +\frac{1}{10}(x-25)\\\\f(x)\approx 5+\frac{1}{10}(x-25)$

We evaluate for x=25.3

$f(25.3)\approx 5+\frac{1}{10}(25.3-25)\\\\f(25.3)\approx 5+\frac{1}{10}(0.3)=5.03$

If we rearrange the approximation to be in the form mx+b we have:

$f(x)\approx 5+\frac{1}{10}(x-25)=5+0.1x-2.5\\\\f(x)\approx 0.1x+2.5$

Then, m=0.1 and b=2.5.