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

zeeskeys

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

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 \(\displaystyle \epsilon\) to ensure that \(\displaystyle f(x)\) is strictly increasing (look at the derivative)?

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

Please feel free to write back if you require further assistance.
 
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