Find the linear approximation of the function
f(x) = sqrt of (1 − x) at a = 0. Use L(x) to approximate the numbers sqrt of 0.9
and sqrt of 0.99.
.

Question

Find the linear approximation of the function 
f(x) = sqrt of (1 − x) at a = 0. Use L(x) to approximate the numbers sqrt of 0.9
 and sqrt of 0.99.
.

anonymous · Answer

I already found L(X) = -1/2X + 1. Idk what do do after that.

myininaya · Answer

$$\sqrt{1-x} \approx \frac{-1}{2}x+1$$


$$\sqrt{0.9}=\sqrt{1-0.1} \approx \frac{-1}{2}(0.1)+1$$

anonymous · Answer

Srry but where did u get sqrt of 1-0.1 from if x = 0.9

myininaya · Answer

.9=1-0.1

myininaya · Answer

I was trying to find out what x was

myininaya · Answer

x is 0.1 since we want to approximate sqrt(0.9)

anonymous · Answer

Ok so for sqrt of 0.99 it wud be $$\sqrt{0.99} = \sqrt{1- 0.01} ~ -1/2 (0.01) + 1$$

myininaya · Answer

You forgot the approximation symbol btw sqrt(1-0.01) and -1/2*(.01)+1
Also you should simplify -1/2*.01+1

anonymous · Answer

so the approximation is 0.995 right?

anonymous · Answer

So what is the general formula to write approxiamtions?

myininaya · Answer

$$\frac{-1}{2} \cdot \frac{1}{100}+1 =\frac{-1}{2} \cdot \frac{1}{100}+\frac{200}{200}=\frac{200-1}{200}=\frac{199}{200} \text{ or } =.995$$
So yes for sqrt(.99), that is a good approximation. 


To write linear approximations:
Say we have a curve y=f(x) and we want to find the linear approximation at x=a for the numbers m.

So we have:

$$\text{ lines have this form: } y=mx+b$$

Well we can find the general slope (aka derivative) for this curve.

$$y'=f'(x)$$

We actually wanted to know the slope at x=a

That slope would in fact by y' evaluated at x=a

So the slope of this tangent line is:

$$y=f'(a)x+b$$
We still need to find the y-intercept, b.

We do know a point on this line (a, f(a))

We can input this point in to find b.

So we have:

$$f(a)=f'(a) \cdot a +b$$
$$f(a)-f'(a) \cdot a =b $$

So the tangent line to y=f(x) at x=a
Or the equation we will use for linear approximations is:

$$y=f'(a)x+f(a)-f'(a) \cdot a $$

or you might prefer to write it as:

$$y=f'(a) \cdot (x-a) +f(a)$$


So anyways this is the linear approximation to y=f(x) for values near x=a.
The approximations will get nastier the further your x is away from a.


So we say:

$$f(x) \approx f'(a)(x-a)+f(a) \text{ for values of x near x }$$

This is the general form for linear approximations.

anonymous · Answer

Are u typing a reply? or is this a glitch?

myininaya · Answer

That slope m and that number m we want to use linear approximations on aren't the same.
lol.

anonymous · Answer

WOW THANK U SOO MUCH FOR SUCH A thorugh explanation :D

myininaya · Answer

Let's say we want to approximate n instead of m.
So we have that 
$$n=f( * ) \approx f'(a)(*-a)+f(a)$$

What we input to approximate n depends on what the function is.

anonymous · Answer

the star * symbol/ what does that represent in tha equation above

anonymous · Answer

is it x?

myininaya · Answer

* is the number I chose to show you that we are not inputting n but we are inputting in a value so that the function is the same as n for some value of x.
Here I just chose that value of x to be *.

anonymous · Answer

Oh I SEE. This makes alot more sense now. Thank u soo much for putting so much time into heloing me with this question . I appreciate it.

myininaya · Answer

Np. I like linear approximations.

anonymous · Answer

Thank u :) I might need help with more just b/c sum I find hard to differentiate

myininaya · Answer

k. Post em separately just in case I'm away. I have chores to do :(.

anonymous · Answer

NO PROBLEM :)