## zeesbrat3 one year ago 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?

1. zeesbrat3

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

2. zeesbrat3

@ganeshie8

3. SolomonZelman

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

4. SolomonZelman

you would need a trig substitution....

5. zeesbrat3

I haven't gotten to integration yet..

6. SolomonZelman

oh, then just recongize the derivative I guess

7. ganeshie8

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

8. SolomonZelman

I feel stupid now:)

9. SolomonZelman

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

10. zeesbrat3

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

11. ganeshie8

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

12. SolomonZelman

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

13. SolomonZelman

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

14. zeesbrat3

So $5 + 2(.03)$

15. ganeshie8

Looks good! it simplifies further tho

16. zeesbrat3

To 5.06

17. ganeshie8

Yep

18. zeesbrat3

Thank you!

19. SolomonZelman

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.

20. ganeshie8

yw

21. SolomonZelman

tnx ganesh.

22. ganeshie8

np