What do I do in this problem?

[Ghost3k]

Ok so my professor gave us some hw, and he did not mention it in his class, but I still want to know how to do it. So can anyone tell me what exactly am I supposed to do in this problem. Not the answers, just the steps that I need to do to get the answers. Here is the question.

Find a linearization function and approximate a.

and b.

The linearization function is

a. The approximation of

is

b. The approximation

is

 

[galactus]

The idea is to use \(\displaystyle f(x_{0}+ \Delta x)\approx f(x_{0})+f'(x_{0})\Delta x\)

Let's start with \(\displaystyle f(x)=\sqrt{x}\)

Then, \(\displaystyle x_{0}=9, \;\ \Delta x=.02\)

\(\displaystyle f'(x)=\frac{1}{2\sqrt{x}}\)

So, using the formula at the top, we get:

\(\displaystyle f(9.02)\approx f(9)+f'(9)(.02)\)

\(\displaystyle \sqrt{9.02}\approx \sqrt{9}+\frac{1}{2\sqrt{9}}(.02)\)

See now what to do?. Now, try it with 8.99

 

[Ghost3k]

The idea is to use \(\displaystyle f(x_{0}+ \Delta x)\approx f(x_{0})+f'(x_{0})\Delta x\)

Let's start with \(\displaystyle f(x)=\sqrt{x}\)

Then, \(\displaystyle x_{0}=9, \;\ \Delta x=.02\)

\(\displaystyle f'(x)=\frac{1}{2\sqrt{x}}\)

So, using the formula at the top, we get:

\(\displaystyle f(9.02)\approx f(9)+f'(9)(.02)\)

\(\displaystyle \sqrt{9.02}\approx \sqrt{9}+\frac{1}{2\sqrt{9}}(.02)\)

See now what to do?. Now, try it with 8.99

thank you, I understand what I need to do.

So essentially it would be
\(\displaystyle \sqrt{8.99}\approx \sqrt{9}+\frac{1}{2\sqrt{9}}(.01)\) correct?