Help with linear approximation please! I need help in part a) my calculation is as follows f(2)+(2.2-2)(f(2)) which then gave me 40+(0.2)(40) which gave me 48 but that is not the correct answer =/ .. Help please!!

Question

anonymous · Answer

attachment

anonymous · Answer

@Luigi0210

luigi0210 · Answer

@Jhannybean @ganeshie8 @primeralph 
Sorry, I threw in the towel awhile ago

luigi0210 · Answer

@johnweldon1993

anonymous · Answer

@e.mccormick help please? =(

e.mccormick · Answer

hmmm... do you know the slope at 2?

e.mccormick · Answer

Cause that graph is a little small for me to read.  Hehe.

anonymous · Answer

I will get it enlarged for ya just one sec

anonymous · Answer

https://webwork.elearning.ubc.ca/webwork2_files/tmp/MATH184-922_2013S1//gif/6TGFHE2ZLS05-1297-setAssignment_10prob3image1.png this help?

anonymous · Answer

slope of 4 at x=2

anonymous · Answer

4*.2 +40
im guessing here

anonymous · Answer

oh what 40.8 is right
what did I do wrong? @_@

e.mccormick · Answer

Yah.  That changes things.

For  linear approximation you use this:
$f(x) \approx f(a)+f'(a)(x-a)$

That pic was the $f'(a)$ part.

anonymous · Answer

and @e.mccormick I thought I did that in my calculations? oO

anonymous · Answer

f'(a) is the first derivative of f(a)

instead of using f ' (a), you ended up using f(a) instead

anonymous · Answer

oo so can someone show me the actual calculation then with numbers please? =/

e.mccormick · Answer

$f(x)=40+(4)(2.2-2)$

anonymous · Answer

ahh darnit ok thanks so much!

anonymous · Answer

4 was determined using the graph

because a=2, we needed to look at the graph for a=2 in order to determine f'(2)

anonymous · Answer

yeah that was cleared up thanks to you guys! Thanks again! The help is much appreciated!

e.mccormick · Answer

Yep, small mistake.  The graph makes you think that is the function, but it was stated as being the prime.  Makes it an easy one to get wrong.