If f(1) = 5 and f prime of x equals the quotient of 6 and the quantity x squared plus 2, which of the following is the best approximation for f(1.03) using local linearization?

Question

zeesbrat3 · Answer

$$f(1) = 5$$ $$f'(x) = \frac{ 6 }{ x^2 + 2}$$

zeesbrat3 · Answer

@ganeshie8

solomonzelman · Answer

you first need to integrate the 6/(x²+2)

solomonzelman · Answer

you would need a trig substitution....

zeesbrat3 · Answer

I haven't gotten to integration yet..

solomonzelman · Answer

oh, then just recongize the derivative I guess

ganeshie8 · Answer

$$f(1+0.03) = f(1) + f'(1)(1.03-1)$$

solomonzelman · Answer

I feel stupid now:)

solomonzelman · Answer

you don't even need to find f(x).

zeesbrat3 · Answer

There is no defined f'(1) I thought?

ganeshie8 · Answer

Since it is an approximation, it makes more sense to use $\approx$

$$f(1+0.03) \approx  f(1) + f'(1)(1.03-1)$$

simplify

solomonzelman · Answer

for f`(1) plug in 1 for x into the f'(x)

solomonzelman · Answer

(and it is always defined, because x²+6 is never=0, it is at least 6)

zeesbrat3 · Answer

So $$5 + 2(.03)$$

ganeshie8 · Answer

Looks good! it simplifies further tho

zeesbrat3 · Answer

To 5.06

ganeshie8 · Answer

Yep

zeesbrat3 · Answer

Thank you!

solomonzelman · Answer

I wish one day I would be that good at math too.... :D
f(1+0.03) was quite awesome, I will remeber this trick.

ganeshie8 · Answer

yw

solomonzelman · Answer

tnx ganesh.

ganeshie8 · Answer

np