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

1. ## 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.

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

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

After that, you should prove that $f(x)=x$ 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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