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Analysis differentiation proof

Let f:[0,∞)→Rf : [0,\infty) \rightarrow \mathbb{R} be continuous and f is differentiable on (0,∞)(0,\infty) and that the lim⁡x→∞f′(x)=1\lim_{x\to \infty} f'(x)=1 prove that there exists a constant M such that f(x)≤M+1.2x, ∀x≥0f(x) \leq M+1.2x, \ \forall x \geq 0

What I did was show the definition for the limit which is exists A ∣f′(x)−1∣<ϵ|f'(x)-1|<\epsilon for x>Ax>A and then I took the definition of mod and set epsilon=0.2making it bounded e.g. f′(x)<1.2f'(x)<1.2 then is it right I can integrate? I did that and got f(x)≤1.2x+cf(x)\leq1.2x+c. My problem with this is if I can integrate that inequality, I believe I can as question says x>0 and this only works for x>A. Any if I'm doing something completely wrong, hints would be appreciated.
Original post by Dilzo999
Let f:[0,∞)→Rf : [0,\infty) \rightarrow \mathbb{R} be continuous and f is differentiable on (0,∞)(0,\infty) and that the lim⁡x→∞f′(x)=1\lim_{x\to \infty} f'(x)=1 prove that there exists a constant M such that f(x)≤M+1.2x, ∀x≥0f(x) \leq M+1.2x, \ \forall x \geq 0

What I did was show the definition for the limit which is exists A ∣f′(x)−1∣<ϵ|f'(x)-1|<\epsilon for x>Ax>A and then I took the definition of mod and set epsilon=0.2making it bounded e.g. f′(x)<1.2f'(x)<1.2 then is it right I can integrate? I did that and got f(x)≤1.2x+cf(x)\leq1.2x+c. My problem with this is if I can integrate that inequality, I believe I can as question says x>0 and this only works for x>A. Any if I'm doing something completely wrong, hints would be appreciated.


If you've got it for x>Ax>A then you've automatically got it for x≥0x \geq 0, because continuous functions on closed bounded intervals are bounded (and attain their bounds).
Original post by Dilzo999
Let f:[0,∞)→Rf : [0,\infty) \rightarrow \mathbb{R} be continuous and f is differentiable on (0,∞)(0,\infty) and that the lim⁡x→∞f′(x)=1\lim_{x\to \infty} f'(x)=1 prove that there exists a constant M such that f(x)≤M+1.2x, ∀x≥0f(x) \leq M+1.2x, \ \forall x \geq 0

What I did was show the definition for the limit which is exists A ∣f′(x)−1∣<ϵ|f'(x)-1|<\epsilon for x>Ax>A and then I took the definition of mod and set epsilon=0.2making it bounded e.g. f′(x)<1.2f'(x)<1.2 then is it right I can integrate? I did that and got f(x)≤1.2x+cf(x)\leq1.2x+c. My problem with this is if I can integrate that inequality, I believe I can as question says x>0 and this only works for x>A. Any if I'm doing something completely wrong, hints would be appreciated.


I don't think you can integrate as you've stated. My first thought was "mean value theorem", which is a much less machinery-heavy way to do it. (After all, you have no guarantee that the integral will even converge.)
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Dilzo999
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Original post by Smaug123
I don't think you can integrate as you've stated. My first thought was "mean value theorem", which is a much less machinery-heavy way to do it. (After all, you have no guarantee that the integral will even converge.)

I'm trying to use the mean value theorem but can't see how, could I use the fact that f'(x)>1/2 eventually then that means f(x) is increasing on some interval?
Original post by Dilzo999
I'm trying to use the mean value theorem but can't see how, could I use the fact that f'(x)>1/2 eventually then that means f(x) is increasing on some interval?


There is no need whatsoever for any machinery. Just the limit definition and linearity of the differential operator. Let g(x)=1.2xg(x)=1.2x

Then consider h=g−fh=g-f. What is lim⁡h′(x)\lim h'(x)? Then use the definition of limit to write an inequality. And then by the observation of the first reply you are done.
(edited 8 years ago)

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