Prove a function > another function. Challenge prob. by me

[lookagain]

This is a challenge problem made by me:


\(\displaystyle Let \ x \ belong \ to \ the \ set \ of \ Real \ numbers.\)

\(\displaystyle For \ every \ Real \ number \ x, \ prove \ the \ following:\)


\(\displaystyle {e} ^{(x^2)} \ + \ \frac{2}{5} \ > \ e^x\)





-------------------------------------------------------------------------------------------------

Note: It is true for x = 0 and for x = 1.

Suggestion:

You could prove part of it for some combination on one of these intervals:


x < 0

0 < x < 1

x > 1

 

[tkhunny]

Re: Prove a function > another function. Challenge prob. b

Series is pretty useful.

\(\displaystyle e^{x} = \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\)

\(\displaystyle e^{x^{2}} = \sum_{n=0}^{\infty}\frac{(x^{2})^{n}}{n!}\)

Subtracting and looking leads to a solid conclusion?

Or move e^x to the left and examine zero, instead. A simple enough first derivative, set to zero shows a single solution around x = 0.631. Second derivative everywhere positive. I think we're done.

What caused you to ponder such a problem?

 

[lookagain]

Re: Prove a function > another function. Challenge prob. b

What caused you to ponder such a problem?

There was a similar problem to this on another website, but I
made a variation to it.