$f(x)=-8e^{-0.6x}+e^{-0.4x}$. Determine maximum concentration of the drug, and the value of it assuming the experiment runs for 12 hours

Question

A person is being tested against a certain drug, and has been injected with it. The drug blood concentration, measured in mg/l, is modeled by the function

$f(x)=-8e^{-0.6x}+e^{-0.4x}$

Here, the variable $x$ measures the time in hours. Determine when does the person have the maximum concentration of the drug in the blood torrent, and the value of it assuming the experiment runs for 12 hours

Can anyone Help me complete this is very confusing. I'm not sure what formula to follow and this question is all over the place

Answer 1

At your behest @Rquin, I will elaborate

$f'(x)=\frac{d}{dx} (-0.8 e^{-0.6x})$$+\frac{d}{dx} ( e^{-0.4x})$

$f'(x)= -0.8 \frac{d}{dx}(e^{-0.6x})+ \frac{d}{dx}(e^{-0.4x})$

$f'(x)= (-0.8)(-0.6)(e^{-0.6x})+ (-0.4)(e^{-0.4x})$

For maxima and minima put f'(x)=0

we get, $(-0.8)(-0.6)(e^{-0.6x})+ (-0.4)(e^{-0.4x})=0$

You will get x=5ln12 which is 12.42

and f'(x)>0 when x $\le12$

So, f(x) will have its maximum value at x=12

f(12)=0.00225706