# exponential question

• Apr 28th 2009, 05:01 PM
mollymcf2009
exponential question
Can someone explain the difference between:

$e^{(x^2)}$ and $e^{x^2}$ (Wondering)

I'm sure it is a simple answer, but as my Calc II final is tomorrow, I'm practically brain dead
Thanks!
• Apr 28th 2009, 05:06 PM
stapel
Other than that $e^{x^2}$ might be taken for $(e^x)^2$ when typed as text, I'm not sure there is a difference. (Wondering)
• Apr 28th 2009, 05:10 PM
mollymcf2009
Quote:

Originally Posted by stapel
Other than that $e^{x^2}$ might be taken for $(e^x)^2$ when typed as text, I'm not sure there is a difference. (Wondering)

Their graphs are different, that is why I ask. Here is how I put them into my calculator:

y = e^(x)^2 *this one looks like your run of the mill exponential compressed vertically

y = e^(x^2) *this one looks like a vertically compressed parabola, vertex @ (0,1)

I just need to know because I'm pretty sure I'll see it tomorrow on my final in the volumes of rotation section :)
• Apr 28th 2009, 05:14 PM
skeeter
Quote:

Originally Posted by mollymcf2009
Their graphs are different, that is why I ask. Here is how I put them into my calculator:

y = e^(x)^2 *this one looks like your run of the mill exponential compressed vertically

your calculator is interpreting this syntax as e^(2x)

y = e^(x^2) *this one looks like a vertically compressed parabola, vertex @ (0,1)

I just need to know because I'm pretty sure I'll see it tomorrow on my final in the volumes of rotation section :)

.
• Apr 28th 2009, 05:20 PM
mollymcf2009
Ok, this is the problem:

Consider the region bounded by the curves:

$y = e^{x^2}$ ........ etc. etc.

So, how do I interpret that? No parentheses anywhere. What does this graph look like?
• Apr 28th 2009, 05:33 PM
mollymcf2009
I think I've got it
Ok, So I'm assuming that since like Skeeter said $e^{(x^2)}$ is being interpreted at $e^{(2x)}$ that the graph of $e^{x^2}$ looks like $e^x$ but compressed vertically is the correct graph.
• Apr 28th 2009, 06:13 PM
skeeter
Quote:

Originally Posted by mollymcf2009
Ok, So I'm assuming that since like Skeeter said $e^{(x^2)}$ is being interpreted at $e^{(2x)}$ that the graph of $e^{x^2}$ looks like $e^x$ but compressed vertically is the correct graph.

that's not what I said ... e^(x)^2 is interpreted as e^(2x) by the calculator, because the calculator sees the expression as the quantity (e^x) squared

e^(x^2) is the correct calculator syntax for the desired even function, $e^{x^2}$.
• Apr 28th 2009, 06:30 PM
mollymcf2009
Quote:

Originally Posted by skeeter
that's not what I said ... e^(x)^2 is interpreted as e^(2x) by the calculator, because the calculator sees the expression as the quantity (e^x) squared

e^(x^2) is the correct calculator syntax for the desired even function, $e^{x^2}$.

Ok, sorry, I guess I just read it wrong. Thanks for clarifying! So the graph DOES look like a compressed parabola. Glad you caught me on that! (Wink)
• Apr 28th 2009, 09:49 PM
Calculus26
Molly

Good Luck on your Final tommorow