the minmum value of# 9

$\begin{aligned} y & = (e^x-2)(e^x -1) \\ & = e^{2x} - 3e^x + 2 \\ & = e^{2x} - 3e^x + \frac 94 - \frac 14 \\ & = \left(e^x - \frac 32\right)^2 - \frac 14 & \small \blue{\text{Since }\left(e^x - \frac 32\right)^2 \ge 0,} \\ & \ge - \frac 14 = \boxed{-0.25} \end{aligned}$

Substitute $z = e^x$ . We are then looking to minimize the value of $z^2 - 3z + 2$ .

Expressing it in vertex form: $\left(z- \displaystyle \frac{3}{2}\right)^2 - \displaystyle \frac{1}{4}$

The minimum value is then $\boxed{\displaystyle -\frac{1}{4}}$ and occurs when $x = \ln \left(\displaystyle \frac{3}{2}\right)$ .

To find the minimum value of a function, we must set it's derivative equal to zero.

$y=(e^{x}-2)(e^{x}-1)$ $\dfrac{\text{d}y}{\text{d}x}=2e^{2x}-3e^{x}$ $2e^{2x}-3e^{x}=0$ $e^{x}=1.5$ $x=\text{ln}(1.5)$

A quick inspection of the graph, a table of nearby values, or the second derivative will confirm that this is where the minimum is. Now, to solve for the minimum, we can substitute $e^{x}=1.5$ back into our initial equation.

$y=(e^{x}-2)(e^{x}-1)$ $y=(1.5-2)(1.5-1)$ $y=(-0.5)(0.5)$ $y=-0.25$

Thus, the minimum value of the function is $\boxed{-0.25}$ .