# Finding critical number for derivative of exponential?

• Nov 23rd 2007, 03:50 PM
emttim84
Finding critical number for derivative of exponential?
Ok, so I have a function that was originally f (x) = e^2x - 3e^x + 2....the problem asks me to find X intercepts, Y intercepts, relative extrema, inflection points, concavity and any horizontal asymptotes...and if you're wondering why all that, it's so I can finally graph it.

So I factored it and determined that it's (e^x - 1) * (e^x - 2), so e^x = 1 and e^x = 2, which means x = ln1 and x = ln2, so there's X-intercepts at (0,0) and (0.693,0). So I got that ok.

Next for the Y-int, I set x equal to 0 and solved for it, and determined e^2(0) - 3e^(0) + 2 = 0 so Y-intercept is also at (0,0).

Now for the relative extrema, this is where I got stuck. I determined the derivative to be f ' (x) = 2e^2x - 3e^x.....and this is where I'm stuck. What would be the best way to factor this out so I can get the critical numbers of the first derivative? If I factor out a e^x, then I have e^x ( 2e^x - 3) which doesn't necessarily help me since if you set the stuff inside parentheses to x=0 then you have ln3/2 but the e^x on the outside, well even if e^(0), it still equals 1...so yeah, I'm probably overlooking something simple, but what are the critical numbers for the first derivative? And more importantly, how did you find them? Thanks guys!
• Nov 23rd 2007, 04:06 PM
galactus
The derivative of said function is

$2e^{2x}-3e^{x}$

$2(e^{x})^{2}-3e^{x}=0$

Let u=e^x:

$2u^{2}-3u=0$

$u(2u-3)=0$

$u=0 \;\ or \;\ \frac{3}{2}$

$e^{x}=0$, undefined.

$e^{x}=\frac{3}{2}, \;\ x=ln(3/2)$
• Nov 23rd 2007, 04:13 PM
emttim84
Quote:

Originally Posted by galactus
The derivative of said function is

$2e^{2x}-3e^{x}$

$2(e^{x})^{2}-3e^{x}=0$

Let u=e^x:

$2u^{2}-3u=0$

$u(2u-3)=0$

$u=0 \;\ or \;\ \frac{3}{2}$

$e^{x}=0, \;\ x=1$

$e^{x}=\frac{3}{2}, \;\ x=ln(3/2)$

Ok, so the critical numbers are $x=ln(3/2)$ as well as $x=1$? Thanks...so then for the second derivative, f " (x) = $4e^{2x}-3e^{x}$ so factoring it, I have $e^{x}(4e^{x}-3)$ so the critical numbers of the second derivative for concavity would be $x=1$ and $x=ln(3/4)$, correct?

Also, am I correct in assuming that there is a horizontal asymptote at $y=2$ due to the fact that a normal exponential function has a H.A. at $y=0$ but the original function has a + 2 constant on the end, thus moving everything (including the H.A.) up by two?
• Nov 23rd 2007, 04:31 PM
galactus
I made a typo. e^1 does not equal 0. I fix. e^x never equals 0.

ln(3/4) is an inflection point, but 1 is not. There is a minimum at x=ln(3/2), but not at x=1.

Also, $\lim_{x\rightarrow{-\infty}}f(x)=2$

Yes, there is a HA at y=2.
• Nov 23rd 2007, 04:34 PM
emttim84
Quote:

Originally Posted by galactus
I made a typo. e^1 does not equal 0. I fix. e^x never equals 0.

ln(3/4) is an inflection point, but 1 is not. There is a minimum at x=ln(3/2), but not at x=1.

Also, $\lim_{x\rightarrow{-\infty}}f(x)=2$

Yes, there is a HA at y=2.

Ok, that makes sense about the e^x thing...thank you so much!