Linear Approximation (using tangent line) of 1/0.104

[chucknorrisfish]

Use linear approximation, i.e. the tangent line, to approximate 1/0.104 as follows: Let f(x)=1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.104 . Then use this to approximate 1/0.104 .

L(x)=f(a)+f'(a)(x-a)...i think.

If f(x)=1/x = x-1 , f'(x) = -x-2.
The "nice" point near 0.104 'might' be .1 since 1/.1 = 10. Now calculate f'(.1) = -.1-2 = -100
Then dx =.004
Then using
L(x)=f(a)+f'(a)(x-a)
L(.104)=f(.1) + f'(.1)*.004
L(.104) =approx. 10 + -100*.004

9.61538 is what i get, and it's not right...So it's totally possible what i did is totally wrong.

 

[pka]

NO! It is correct if you have stated the problem correctly.
1/(0.104) is about 9.615.
But is the problem 1/104 ?

 

[chucknorrisfish]

no it says .104

 

[chucknorrisfish]

Oh, it is right.

9.615 is right. The thing just wanted 9.6, so that's why it wasnt working. thanks tho