# Math Help - Help understanding max and min with log derivatives

1. ## Help understanding max and min with log derivatives

Hello, I'm having trouble with using the max and min algorithm (finding the derivative, setting it to 0, and isolating) to find absolute max and min values for log and exponential functions. My teacher mentioned something about taking the ln of both sides when you have e, and taking the e of both sides when you have ln- however, I'm confused as to how to do this. For example, this question:

Determine the absolute max and min values for this function: (e^-x) - (e^-3x), where x is less than or equal to 10, and greater than or equal to 0.

I found the derivative to be -e^-x + 3e^-3x. Now I know I need to set it 0, and solve for x- however I'm thrown off by the e base. How do I solve for x when I have e? And vice versa, if I had ln, how would I solve for x?

I'd greatly appreciate any help, especially as I have to know this for a test tomorrow. Thank you in advance!

2. You have the correct derivative. One thing you can do is let $u=e^{-x}$

Then you have $3u^{3}-u=0$

Factor: $u(3u^{2}-1)=0$

Now, solve for u and remember that $u=e^{-x}$

3. Hello, starswept!

You need to brush up on your PreCalculus . . .

Determine the absolute max and min values for:
. . $y \;=\;e^{-x} - e^{-3x},\;\;0 \leq x \leq 10$

I found the derivative to be: . $y' \:=\: -e^{-x} + 3e^{-3x}$ . . . . Good!

We have: . $-e^{-x} + 3e^{-3x} \;=\;0$

Multiply by $e^{3x}\!:\;\;-e^{2x} + 3 \;=\;0\quad\Rightarrow\quad e^{2x} \:=\:3$

Take the log of both sides: . $\ln\left(e^{2x}\right) \:=\:\ln3\quad\Rightarrow\quad 2x\!\cdot\!\ln e \:=\:\ln3$

. . and we have: . $2x \:=\:\ln 3\quad\Rightarrow\quad\boxed{x \,=\,\frac{1}{2}\ln3}$ .
. . . a critical value.

And: . $y \;=\;e^{-\frac{1}{2}\ln3} - e^{-3\cdot\frac{1}{2}\ln3}\quad\Rightarrow\quad y \:=\:\frac{2\sqrt{3}}{9} \:\approx\:0.385$

The second derivative is: . $y'' \;=\;e^{-x} - 9e^{-3x} \;=\;\frac{1}{e^x} - \frac{9}{e^{3x}} \;=\;\frac{e^{2x} - 9}{e^{3x}}$

When $x \,=\,\frac{1}{2}\ln3\!:\;\;y'' \;=\;\frac{e^{2\cdot\frac{1}{2}\ln3} - 9}{e^{3\cdot\frac{1}{2}\ln3}} \;=\;\frac{e^{\ln3} - 9}{e^{\frac{3}{2}\ln 3}} \;=\;\frac{3 - 9}{e^{\ln3^{\frac{3}{2}}}} \;=\;\frac{-6}{3^{\frac{3}{2}}}$ . . . .
negative

. . Hence, the graph is concave down; we have a maximum point.

We must also examine the endpoints of the interval.

At $x = 0\!:\;\;y \:=\:e^0 - e^0 \:=\:0$

At $x = 10\!:\;\;y \:=\:e^{-10} - e^{-30} \:=\:0.0000454$

Therefore, . $\begin{Bmatrix}\text{absolute maximum:} & \left(\frac{1}{2}\ln3,\,\frac{2\sqrt{3}}{9}\right) \\
\text{absolute minimum:} & (0,\,0)\end{Bmatrix}$