Results 1 to 4 of 4

Math Help - Prove

  1. #1
    Junior Member
    Joined
    Jan 2010
    From
    TAIWAN
    Posts
    39

    Prove

    show the function has an inverse
    f(x)=x^{3}-x+1

    the ft maybe do not have inverse
    so
    if have ,prove it!

    my idea is use strictly insresing(decresing) to show
    but i do not know the next step to show
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member mathemagister's Avatar
    Joined
    Feb 2010
    Posts
    191
    Quote Originally Posted by chialin4 View Post
    show the function has an inverse
    f(x)=x^{3}-x+1

    the ft maybe do not have inverse
    so
    if have ,prove it!

    my idea is use strictly insresing(decresing) to show
    but i do not know the next step to show
    Your idea would work fine. If you can prove that f(x) is always increasing or decreasing, you have proven that f(x) has an inverse.

    f(x) increases when f'(x)>0
    f(x) decreases when  f'(x)<0

    f'(x) = 3x^2 -1

    Now we need to check if f'(x) changes sign (crosses the x-axis). If it does change sign, the f(x) does not have an inverse. First, we check for zeros of the derivative:

    3x^2 - 1 = 0
    x^2 = \frac{1}{3}
    Zeros of f'(x) are:
    x=\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}

    This does not conclusively prove anything though, because it only says that the derivative has zeros, but not that it changes sign. To check if it changes sign, just figure out if the derivative is positive or negative in the regions between the zeros:

    f'(0)=-1 [substitute 0 into the derivative equation] Since the derivative is negative, f(x) is decreasing at x=0.
    f'(1)=2 Since the derivative is positive, f(x) is increasing at x=1.

    This proves that f(x) does not have an inverse because it is increasing at some point, but decreasing at another.

    Hope I helped
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    From
    TAIWAN
    Posts
    39
    Quote Originally Posted by mathemagister View Post
    Your idea would work fine. If you can prove that f(x) is always increasing or decreasing, you have proven that f(x) has an inverse.

    f(x) increases when f'(x)>0
    f(x) decreases when  f'(x)<0

    f'(x) = 3x^2 -1

    Now we need to check if f'(x) changes sign (crosses the x-axis). If it does change sign, the f(x) does not have an inverse. First, we check for zeros of the derivative:

    3x^2 - 1 = 0
    x^2 = \frac{1}{3}
    Zeros of f'(x) are:
    x=\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}

    This does not conclusively prove anything though, because it only says that the derivative has zeros, but not that it changes sign. To check if it changes sign, just figure out if the derivative is positive or negative in the regions between the zeros:

    f'(0)=-1 [substitute 0 into the derivative equation] Since the derivative is negative, f(x) is decreasing at x=0.
    f'(1)=2 Since the derivative is positive, f(x) is increasing at x=1.

    This proves that f(x) does not have an inverse because it is increasing at some point, but decreasing at another.

    Hope I helped

    thx for ur help!!! such a nice guy!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member mathemagister's Avatar
    Joined
    Feb 2010
    Posts
    191
    Quote Originally Posted by chialin4 View Post
    thx for ur help!!! such a nice guy!!
    You're very welcome
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 10th 2010, 11:28 AM
  2. Prove(10)
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: February 26th 2010, 04:16 AM
  3. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  4. How do I prove b^x * b^y = b ^ x + y on R?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 25th 2008, 12:15 AM
  5. Prove:
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 24th 2008, 11:45 PM

Search Tags


/mathhelpforum @mathhelpforum