A word problem gives this cost equation and asks to find the x where the average cost is minimized, to do so, I need to solve for average cost and derive it and then set it to zero and solve;
$c(x) = x^2+10xe^{-x}$, cost in relation to quantity
$AC = \frac{c(x)}{x}$, average cost
$\frac{d(ac)}{dx} = 1+(-x10e^{-x})$, derivative of average cost, I think
$0 = 1-x10e^{-x}$ ?
here is where I'm really lost, I don't know how to solve for ZERO when there is $e^{-x}$ in the equation.
Can anyone please help me? I can't seem to find any place that can help me with this.
Your derivative for the average cost is incorrect.
$\frac{d(ac)}{dx} = 1-10e^{-x}$
Now,
$$0 = 1 - 10e^{-x}\\$$
$$e^{-x} = \frac{1}{10} ~\text{(dividing by 10)}\\$$
$$e^{-x} = \frac{1}{10}\\$$
$$-x = -2.3 ~\text{(taking the ln of both sides)}$$
Therefore, $x = 2.3$.