If $f'(x) = -f(x)$ and $f(1)=1$, then what is $f(x)$? If $f'(x) = -f(x)$ and $f(1)=1$, then $f(x)=$
(a) $1/2e^{-2x+2}$
(b) $e^{-x-1}$
(c) $e^{1-x}$
(d) $e^{-x}$
(e) $-e^{x}$
If you plug in $1$ for $x$ in option c it gives you $e^{1-1}$ or $e^0$ which is $1$ hence what is needed.
 A: Hint: All you need to do in this case is to evaluate $f(1)$: $f_A(1), f_B(1), f_C(1), f_D(1), f_E(1)$.
Show your calculations in each case.
Doing that rules out all possibilities except $(C)$, irregardless of the first condition, though you want to evaluate $f'_C(x)$ to show that $f_C = f'_C$.
A: If $f'(x) = -f(x)$,
then
$0 = f'(x)+f(x)
= e^x(f'(x)+f(x))
= (e^x f(x))'
$
so $e^x f(x)$ is constant.
Set $e^x f(x) = c$
and set $x = 1$.
Then, since $f(1) = 1$,
$c = e f(1) = e$
so $f(x) = e e^{-x} = e^{1-x}$.
An alternative solution is to write
$f'(x)/f(x) = -1$
so $(\ln(f(x))'=-1$
so $\ln f(x) = c-x$
or $f(x) = e^{c-x}$
for some constant $c$.
This has the problem that $f(x) = 0$ causes problems,
and the first solution does not have this problem.
A: This is a separable differential equation.  We have $\frac{dy}{dx} = -y$, so
$$\int \frac{dy}{y} = \int -dx$$
and we get get the general solution $y = e^{-x + k}$.  The initial condition $y(1) = 1$ implies $k = 1$ in this case.
A: If $f'(x) = -f(x)$,
then
$\frac{f'(x)}{f(x)}=-1 \implies \int \frac{f'(x)}{f(x)}dx=-\int dx \implies f(x)= e^{-(x-k)}\ $ (Here k is the integration constant) 
so $e^x f(x)$ is constant.
Here, $e^x f(x) = e^k$
and set $x = 1$.
Then, since $f(1) = 1$,
$e f(1) =e^k $
so $f(x) = e e^{-x} = e^{1-x}$.
so, $f(x)= e^{1-x} \implies\ \ $ Option (b) is correct. 
