Thread: Finding critical number for derivative of exponential?

1. 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!

2. The derivative of said function is

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

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

Let u=e^x:

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

$\displaystyle u(2u-3)=0$

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

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

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

3. Originally Posted by galactus
The derivative of said function is

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

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

Let u=e^x:

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

$\displaystyle u(2u-3)=0$

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

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

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

Also, am I correct in assuming that there is a horizontal asymptote at $\displaystyle y=2$ due to the fact that a normal exponential function has a H.A. at $\displaystyle y=0$ but the original function has a + 2 constant on the end, thus moving everything (including the H.A.) up by two?

4. 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, $\displaystyle \lim_{x\rightarrow{-\infty}}f(x)=2$

Yes, there is a HA at y=2.

5. 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, $\displaystyle \lim_{x\rightarrow{-\infty}}f(x)=2$

Yes, there is a HA at y=2.
Ok, that makes sense about the e^x thing...thank you so much!