anonymous
  • anonymous
Use linear approximation, i.e. the tangent line, to approximate 999^4 as follows: Let f(x)=x^4. The equation of the tangent line to f(x) at x=10^3 is best written in the form y=f(a)+f'(a)*(x−a) where a=10^3, f(a)=(10^3)^4, and f'(a)=4(10^3)^3. Using this, we find our approximation: 999^4 approximately equals? ______ I'm stuck at the last part and can't figure it out. Can someone walk me through it without giving me the answer?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmate
  • mathmate
Straight substitution of your formula gives: f(x)=x^4 f'(x)=4x^3 let a=1000, f(x)=f(a)+f'(x)(x-a) =1000^4 +4(1000^3)(999-1000) =996000000000 compare with exact value of 999^4 =996005996001
anonymous
  • anonymous
OOOHHH!! So I plug 999^4 into x in the equation!
mathmate
  • mathmate
You probably made a miscalculation for f'(a)=4(10^3)^3 with a ^3 too many.

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mathmate
  • mathmate
The only time you use 999 is (999-1000). You do not need 999 for f(a) nor f'(a), since a=1000.
mathmate
  • mathmate
Sorry, I had a typo above for the equation, but the calculation that followed is correct. should read: f(x)=f(a)+f'(a)(x-a)
anonymous
  • anonymous
Why do I use 999 instead of 999^4 at the end of the equation?
anonymous
  • anonymous
I did not, I put in 999^4 as x in 1000^4 +4(1000^3)(x-1000) and got the wrong answer. I put in 999 for x and got the right answer.
anonymous
  • anonymous
I think it's because 999 is x and f(x)=x^4, which is where the rest of my equations come from.
anonymous
  • anonymous
I put that in and the program counted it as wrong, make 999^4 just 999 and it counts it right.
anonymous
  • anonymous
Dude, I don't know, I copied and pasted the question and put in the correct symbols where they needed to be. I was surprised I made it to the end without help. If the program says 999 is x, instead of 999^4, and I get the problem right I won't argue. I'm sure it was just the wording of the problem that threw us off.
mathmate
  • mathmate
Sorry I was offline. (999-1000) represents delta-X in the differential, that's why it is not raised to the fourth power. The equation that you correctly stated was: f(x)=f(a)+f'(a)(x-a) where a=1000, (x-a) = dx, so the linear approximation becomes: f(x)=f(a)+f'(a)[delta-x] which geometrically is shown in the figure below:|dw:1326861483536:dw|

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