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Find Point c that satisfies the Mean Value Theorem

  1. Apr 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the point "c" that satisfies the Mean Value Theorem For Derivatives for the function
    ## f(x) = \frac {x-1} {x+1}## on the interval [4,5].
    Answer - c = 4.48
    2. Relevant equations
    ##x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}##
    ##f'(c) = \frac { f(b) - f(a)} {b-a}##

    3. The attempt at a solution
    I found the derivative
    ##f'(x) = \frac {2} {(x+1)^2}##
    a) ## f(4) = \frac {4-1} {4+1} = \frac {3} {5}##
    b) ## f(5) = \frac {5-1} {5+1} = \frac {2} {3}##
    Substituting in my values
    ## \frac {2} {(x+1)^2} = \frac { \frac {2} {3} - \frac {3} {5}} {5-4}##
    ## \frac {2} {(x+1)^2} = \frac {1} {15}##
    ## 30 = c^2 +2c +1##
    ## 0 = c^2 +2c -29##
    ##x = \frac {-2 \pm \sqrt{2^2 -(4)(1)(-29)}} {2(1)}##
    Which gives me 3.475 and -7.475. I'm not sure where I went wrong, any help would be greatly appreciated. Neither one of there are in my interval and my solution says it needs to be 4.48.
     
  2. jcsd
  3. Apr 6, 2017 #2
    look the equation:
    $$x = \frac {-2 \pm \sqrt{2^2 -(4)(1)(-29)}} {2(1)}$$
    and solve it carefully
     
  4. Apr 6, 2017 #3
    I swear I put that in my calculator a few times but this last time it worked... Thanks a lot anyways :)
     
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