Science:Math Exam Resources/Courses/MATH110/April 2013/Question 02 (a)
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q3 (c) • Q3 (d) • Q3 (e) • Q4 • Q5 • Q6 • Q7 • Q8 • Q9 (a) • Q9 (b) •
Question 02 (a) |
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Sketch a single function satisfying all of the listed criteria. |
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 |
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Think about what each condition means. The first is asking for a function value at 3 to be equal to 0. What does the sign of the first and second derivative mean for the shape of the function? |
Hint 2 |
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The first derivative tells us if the function is increasing or decreasing. Since the second condition says that that the first derivative is negative at 3, function is decreasing at 3. |
Hint 3 |
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The second derivative tells us if the function is concave up or concave down. Since the third condition is saying that the second derivative is positive at 3, the function is concave up at 3. |
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.
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Solution |
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Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. A simple-like parabola with a zero at three will work. The other zero one can be anywhere, for no particular reason lets make it at 5. So the function satisfies all three conditions. To double check,
Let's see the picture Note: Many other functions are possible as well, it does not even have to be a second order polynomial. |