Science:Math Exam Resources/Courses/MATH110/December 2011/Question 09

MATH110 December 2011

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Question 09
Write down an algebraic expression for a function $f$ satisfying the following four criteria: $\lim _{x\to 0^{-}}f(x)=2$ $\lim _{x\to 0^{+}}f(x)=-1$ $\lim _{x\to -\infty }f(x)=0$ $\displaystyle f'(x)=-2{\hbox{ for }}x>0$ Then sketch the function.

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
Try drawing a picture of what the graph of this function would look like. Are there some easy functions that you can piece together to satisfy these properties?

Hint 2
Which function $f(x)$ has the property that its derivative $f'(x)=-2$?

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. The best way to approach this function is as a piecewise function. The statement of the problem suggests that there will need to be at least two pieces to the final function: one for $x>0$ and one for $x<0$. We will start by considering the $x>0$ piece. We know for $x>0$ that $f'(x)=-2$. The only function that has -2 as its derivative is the line with slope -2. So we know that this piece of our function will look something like $y=-2x+b$. Looking at the second requirement of the function, we know that the limit of this function as it approaches 0 must be -1. Thus our line should intersect the y-axis at -1 and one piece of our function will be y = -2x - 1. Now we consider the piece for $x<0$. We know that the limit approaching x = 0 must be 2 and that the limit to negative infinity must be 0. There are many functions that satisfy the first condition; an easy example is y = 2. There are also lots of functions satisfying the second condition; two examples are $e^{x}$ and ${\frac {1}{x}}$. There are multiple ways to continue. We could break this piece of the function into two more pieces, using the examples I've cited above to get the following piecewise: $\displaystyle f(x)={\begin{cases}e^{x}&x\leq -1\\2&-1<x\leq 0\\-2x-1&x>0\end{cases}}$ A more complex solution would be to take a function that with the appropriate limit as x goes to negative infinity, like $y=-{\frac {1}{x}}$ and shift it so that its y-intercept is 2. This would produce a piecewise like this: $\displaystyle f(x)={\begin{cases}-{\frac {1}{x-1/2}}&x\leq 0\\-2x-1&x>0\end{cases}}$

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