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, $(G(H(x)){)}^{\prime }=G^{\prime }(H(x))H^{\prime }(x)$, where $G(x)=e^x$ and $H(x)=3-x$. Then $G^{\prime }(x)=e^x$ and $H^{\prime }(x)=-1$, so we calculate:

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

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,

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

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.