Science:Math Exam Resources/Courses/MATH110/April 2018/Question 02 (d)

MATH110 April 2018

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  Q7 (e)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  Q9  •  Q10  •  

   Other MATH110 Exams

  •  December 2011  •  December 2010  •  April 2012  •  April 2011  •  December 2012  •  April 2013  •  December 2013  •  April 2014  •  December 2014  •  December 2015  •  April 2017  •  April 2016  •  December 2016  •  December 2017  •  April 2018  •  

Question 02 (d)

Consider a function ${\textstyle f}$ whose graph is showed below. On which of the following intervals does the Mean Value Theorem (MVT) apply to ${\textstyle f}$? Select all that apply. If the theorem applies, give a reasonable estimate of the number ${\textstyle c}$ in the given interval that satisfies the conclusion of the theorem. (a) ${\textstyle [-1.5,0]}$, and the number ${\textstyle c}$ that satisfies the conclusion of MVT is ____ (b) ${\textstyle [-1,0]}$, and the number ${\textstyle c}$ that satisfies the conclusion of MVT is ____ (c) ${\textstyle [-1,1]}$, and the number ${\textstyle c}$ that satisfies the conclusion of MVT is ____ (d) ${\textstyle [0,2]}$, and the number ${\textstyle c}$ that satisfies the conclusion of MVT is ____

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
For the MVT to apply on an interval ${\textstyle (a,b)}$, the function must be continuous on the closed interval ${\textstyle [a,b]}$ and differentiable on the open interval ${\textstyle (a,b)}$.

Hint 2
The number ${\textstyle c}$ is the ${\textstyle x}$-coordinate at which the average slope of ${\textstyle f(x)}$ on the interval ${\textstyle (a,b)}$ is attained.

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 MVT does not apply for (a), since the function is undefined (and thus not continuous) at the point ${\textstyle x=-1.5}$. The MVT does apply for (b), since the function is continuous on the closed interval $[-1,0]$ and differentiable on the open interval $(-1,0)$. The average slope of ${\textstyle f}$ on this interval is ${\textstyle {\frac {f(0)-f(-1)}{0-(-1)}}={\frac {0-1}{1}}=-1}$, so a reasonable estimate for ${\textstyle c}$ would be something around ${\textstyle -0.5}$ since the tangent line to ${\textstyle f}$ at ${\textstyle x=-0.5}$ seems to have slope around ${\textstyle -1}$. The MVT does apply for (c), since the function is continuous on the closed interval $[-1,1]$ and differentiable on the open interval $(-1,1)$. The average slope of ${\textstyle f}$ on this interval is ${\textstyle {\frac {f(1)-f(-1)}{1-(-1)}}={\frac {1-1}{2}}=0}$, so a reasonable estimate for ${\textstyle c}$ would be ${\textstyle 0}$ since the tangent line to ${\textstyle f}$ at ${\textstyle x=0}$ seems to have slope equal to ${\textstyle 0}$. The MVT does not apply for (d), since the function is not differentiable at the point ${\textstyle x=1}$ which lies within the open interval $(0,2)$. Answer: The correct answers are ${\textstyle {\color {blue}(b){\text{ (with }}c\approx -0.5{\text{) and }}(c){\text{ (with }}c\approx 0{\text{)}}}}$.