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Use the Mean Value Theorem to prove that f'(x) = 0.01 has a solution in the (-4, -3) interval. Given (picture attached as well): f(x) = (x+3)^3 / (4(x-1)^2 ) = 12 / ((x-1)) + 11/4 + 16 / (x-1)^2 + x/4 f′(x)=((x+3)^2 (x-9)) / (4(x-1)^3) f′′(x)=(24(x+3)) / (x-1)^4

Use the Mean Value Theorem to prove that f'(x) = 0.01 has a solution in the (-4, -3) interval. Given (picture attached as well): f(x) = (x+3)^3 / (4(x-1)^2 ) = 12 / ((x-1)) + 11/4 + 16 / (x-1)^2 + x/4 f′(x)=((x+3)^2 (x-9)) / (4(x-1)^3) f′′(x)=(24(x+3)) / (x-1)^4

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.3: Algebraic Expressions
Problem 40E
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Use the Mean Value Theorem to prove that f'(x) = 0.01 has a solution in the (-4, -3) interval.

Given (picture attached as well):

  • f(x) = (x+3)^3 / (4(x-1)^2 ) = 12 / ((x-1)) + 11/4 + 16 / (x-1)^2 + x/4
  • f′(x)=((x+3)^2 (x-9)) / (4(x-1)^3)
  • f′′(x)=(24(x+3)) / (x-1)^4 

 

(x+ 3)3 12 11 + + 16 X f(x) + (х — 1)2 4 - 4(х — 1)2 (х — 1) " 4 (x+ 3)²(x – 9) - 24(x+3) | f'(x) = f"(x) = 4(x – 1)3 (х — 1)4 - |
Transcribed Image Text:(x+ 3)3 12 11 + + 16 X f(x) + (х — 1)2 4 - 4(х — 1)2 (х — 1) " 4 (x+ 3)²(x – 9) - 24(x+3) | f'(x) = f"(x) = 4(x – 1)3 (х — 1)4 - |
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