use differential to approximate change in f(x)

[PaulKraemer]

Hi,

I am getting a different answer than the back of my book on the following problem:

Use differentials to approximate the change in f(x) = 4x^5 - 6x^4 + 3x^2 - 5 if x changes from 1 to 1.03

My work:

f(x) = 4x^5 - 6x^4 + 3x^2 - 5
dy = (20x^4 - 24x^3 + 9x)dx
when x = 1 and dx = 1.03 - 1 = 0.03, this gives me:
dy = (20 - 24 + 9) (0.03) = 0.15

The back of the book says the answer is 0.06. Did I do something wrong or is the back of the book wrong?

Any help will be greatly appreciated.

Thanks in advance,
Paul

 

[galactus]

You merely have a 9 where a 6 belongs.

\(\displaystyle dy=(20x^{4}-24x^{3}+\underbrace{6x}_{\text{here}})dx\)

\(\displaystyle dx=.03\)

So, with x=1, \(\displaystyle dy=(20-24+6)(.03)=.06\)

 

[PaulKraemer]

Thank you Galactus - I appreciate your help!