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

MATH110 April 2012

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Question 09
An antiderivative of a function ƒ is a function F satisfying $\displaystyle F'(x)=f(x)$ for all x. Find two antiderivatives of the function $\displaystyle e^{3-x}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that $e^{x}$ is its own derivative.

Hint 2
Suppose you have one function $f(x)$ that has a derivative of $g(x)$. Can you think of an easy function $h(x)$ that has derivative zero so that when you add to $f(x)$ you get $\displaystyle (f(x)+h(x))'=f'(x)+h'(x)=f'(x)+0=g(x)$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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