Science:Math Exam Resources/Courses/MATH100/December 2012/Question 10

MATH100 December 2012

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Question 10
Full-Solution Problems. In questions 5-11, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. Find a function $f(x)$ such that $\displaystyle f'(x)=x^{3}$ and such that the line $x+y=0$ is tangent to the graph of $y=f(x)$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Antidifferentiating gives you step one. However remember that antidifferentiating only gives you the answer up to a constant. We need to use the other information given to help figure our constant. How do we relate slopes and derivatives?

Checking a solution serves two purposes: helping you if, after having used the hint, 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. To find the form of the function we compute the antiderivative, $f(x)={\frac {x^{4}}{4}}+C.$ To find the actual function we are interested in, we need to determine C. The extra information we are given is that we want the line $\displaystyle x+y=0$, which is equivalent to $\displaystyle y=-x$ to be tangent at some point on the function $\displaystyle {}f(x)$. The slope of this line is -1 and thus we need $\displaystyle f'(x)=x^{3}=-1$ which happens when $\displaystyle {}x=-1$. Plugging this point into the line we have that $\displaystyle {}y=-(-1)=1$ and so since the line is tangent to the function there, $\displaystyle {}f(-1)=1$ as well. Therefore we have, ${\begin{aligned}f(-1)&={\frac {1}{4}}+C=1\\C&=1-{\frac {1}{4}}={\frac {3}{4}}.\end{aligned}}$ Thus, our desired function is $\displaystyle y={\frac {x^{4}}{4}}+{\frac {3}{4}}$.