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Thread: Deriving Function: Suppose g is real-valued differentiable fcn, |g'(x)| <= M

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    Deriving Function: Suppose g is real-valued differentiable fcn, |g'(x)| <= M

    Suppose that g is a real valued, differentiable function whose derivative g' satisfies the inequality |g'(x)|less than or equal to M for all x in R.
    Show that if epsilon is greater than 0 is small enough, then the real valued function f defined by f(x)=x+epsilon*g(x) is one to one and onto.
    Recall that a function f is said to be "one to one" if x sub 1 does not equal x sub 2 implies that f(x sub 1) does not equal f(x sub 2), and f is said to be "onto" if for every real number y, there is a real number x such that f(x) = y.

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    Quote Originally Posted by zeeskeys View Post
    Suppose that g is a real valued, differentiable function whose derivative g' satisfies the inequality |g'(x)|less than or equal to M for all x in R.
    Show that if epsilon is greater than 0 is small enough, then the real valued function f defined by f(x)=x+epsilon*g(x) is one to one and onto.
    Recall that a function f is said to be "one to one" if x sub 1 does not equal x sub 2 implies that f(x sub 1) does not equal f(x sub 2), and f is said to be "onto" if for every real number y, there is a real number x such that f(x) = y.
    Hi zeeskeys,

    If you could take [tex]\epsilon=0[/tex], then [tex]f(x)=x[/tex] would be one-to-one and onto. In essence, you must prove that, by taking [tex]\epsilon[/tex] small enough, you can make [tex]f(x)[/tex] not too different from [tex]x[/tex].

    As you should know, a strictly increasing, unbounded, and continuous function is one-to-one and onto (write back if you don't know how to prove that). Can you find how to choose [tex]\epsilon[/tex] to ensure that [tex]f(x)[/tex] is strictly increasing (look at the derivative)?

    After that, you should prove that [tex]f(x)=x[/tex] is continuous (easy) and unbounded; this requires looking at the derivative again.

    Please feel free to write back if you require further assistance.
    Last edited by barrick; 11-15-2017 at 04:56 AM. Reason: Correction

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