Science:Math Exam Resources/Courses/MATH110/April 2012/Question 09/Solution 1

Because $e^{x}$ is its own derivative, our anti-derivative for the function $e^{3-x}$ will probably be very similar to the original function. To check our first anti-derivate candidate $e^{3-x}$ we have to use the chain rule, $(G(H(x)))'=G'(H(x))H'(x)$, where $G(x)=e^{x}$ and $H(x)=3-x$. Then $G'(x)=e^{x}$ and $H'(x)=-1$, so we calculate:

$\displaystyle (e^{3-x})'=(G(H(x)))'=G'(H(x))H'(x)=(e^{3-x})(-1)=-e^{3-x}.$

Hence our candidate almost works, we just have to bring the minus sign over: $F(x)=-e^{3-x}$. Using the chain rule to double check, we find that, indeed,

$\displaystyle F'(x)=(-e^{3-x})'=(-e^{3-x})(-1)=e^{3-x}=f(x).$

This is one anti-derivative. To find another one, we can simply add a constant to the anti-derivative shown above, say $F_{2}(x)=-e^{3-x}+5$. When we differentiate, the constant will disappear, giving us the same derivative as before.