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Hi just have a quick question and need some guidance,
Use calculus to show that x<=e^x (for any x element of R)
Hi samuel12;
For x < 0 it is obvious, the LHS is negative and the RHS is positive:
For x>=0, Taylorize e^x. The Taylor series for e^x converges for all x ∈ R.
Subtract x from both sides:
which is obviously true.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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'Use calculus'
When x = 0, e^x = 1 which is greater.
d(x)/dx = 1 for all x
d(e^x)/dx = e^x > 0 for all x => y = e^x is an increasing function.(always goes up never down)
So y = e^x starts bigger and goes on getting bigger (at a faster rate) so y = x will never overtake it.
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thanks for the help
Hi samuel12;
Welcome to the forum!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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