$e^{\large\frac{-t}{5}}(1-\frac{t}{5})=0$ solve for t I'm given the following equation an i have to solve for $"t"$ This is actually the derivative of a function, set equal to zero:
$$f'(t) = e^\frac{-t}{5}(1-\frac{t}{5})=0$$
I will admit im just stuck and im looking for how to solve this efficiently.
steps $1,\;2$ - rewrote the equation and distributed:
$$\frac{1}{e^\frac{t}{5}}(1-\frac{t}{5}) \iff \frac{1-\frac{t}{5}}{e^\frac{t}{5}} $$
steps $3, \;4$ - common denominator of 5,  multiply by reciprocal of denominator:
$$ \frac{\frac{5-t}{5}}{e^\frac{t}{5}}\iff\frac{\frac{5-t}{5}}*\frac{1}{e^\frac{t}{5}}  = \frac{5-t}{5e^\frac{t}{5}}  $$
step 5, set  this is where I,m stuck:
$$f'(t) = \frac{5-t}{5e^\frac{t}{5}}=0$$
How do i go from here? And am I even doing this correctly? Any help would be greatly appreciated.
Miguel
 A: Hint: 
Given either presentation of your derivative: $$e^{-t/5}(1-\frac{t}{5})=0\tag{1}$$
$$\iff \frac{5-t}{5e^{t/5}}=0\tag{2}$$
Note that neither the factor $\,e^{-t/5}\,$ in $(1)$, nor the denominator $\,e^{t/5}\,$ of $(2)$ will ever evaluate to zero: i.e. $e^{-t/5} = \dfrac 1{e^{t/5}}\neq 0,\;$ whatever the value of $t$.
With this in mind, for simplicity, we'll look at equation $(2)$: your derivative will equal zero when and only when the numerator equals zero.
I.e., solve for $t$ given: $\quad 5 - t = 0$.
A: $e^{\dfrac{-t}{5}}\neq0$. (Cause: $\frac{1}{e^\frac{t}{5}}$ is a real number)
Therefore, the other term has to be zero.
Now you can solve it !
A: Here's an even simpler approach. Your equation is a product of terms, so one of them (or both) must equal $0$ in order for the product to equal $0$.
From the first term one gets $e^{-\frac{t}{5}} = 0$ which is asymptotically true as $t \rightarrow \infty$.
From the second term one gets $1 - \frac{t}{5} = 0$ which has the obvious solution $t=5$.
