Prove that f is strictly increasing

• Jan 18th 2010, 08:13 AM
feyomi
Prove that f is strictly increasing
f:[0,∞) → ℝ, f(x) = e^(x^2)

In order to prove that f is strictly increasing, do I have to show that f'(x) is always positive?

I think f'(x)=[(x^2)e^(x^2)]/e + 2xe^(x^2)lne
I could have gotten that horribly wrong, so correct me if I have.

Then how to prove that this is strictly positive?

Thanks.
• Jan 18th 2010, 08:22 AM
feyomi
Quote:

Originally Posted by feyomi
f:[0,∞) → ℝ, f(x) = e^(x^2)

In order to prove that f is strictly increasing, do I have to show that f'(x) is always positive?

I think f'(x)=[(x^2)e^(x^2)]/e + 2xe^(x^2)lne
I could have gotten that horribly wrong, so correct me if I have.

Then how to prove that this is strictly positive?

Thanks.

Sorry just to add to that. Is it enough to say that each term in f'(x) is positive since xϵ[0,∞) and since e>0 and so f'(x)>0, thus f is strictly increasing?

Thanks
• Jan 18th 2010, 08:26 AM
skeeter
Quote:

Originally Posted by feyomi
f:[0,∞) → ℝ, f(x) = e^(x^2)

In order to prove that f is strictly increasing, do I have to show that f'(x) is always positive?

I think f'(x)=[(x^2)e^(x^2)]/e + 2xe^(x^2)lne
I could have gotten that horribly wrong, so correct me if I have.

Then how to prove that this is strictly positive?

Thanks.

\$\displaystyle f(x) = e^{x^2}\$

\$\displaystyle f'(x) = 2x e^{x^2}\$

for \$\displaystyle x > 0\$ , \$\displaystyle f'(x) > 0\$
• Jan 18th 2010, 08:29 AM
feyomi
Hahaha of course. Silly me.
Thank you.