Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (l)

MATH100 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q5 (e)  •  Q6  •  Q7  •  Q8  •  

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Question 01 (l)

Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question. If ƒ(-1) = 9 and ƒ'(x) ≥ 3 for -1 ≤ x ≤ 2, how small can ƒ(2) possibly be?

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
What does ƒ'(x) ≥ 3 for -1 ≤ x ≤ 2 mean in terms of the increasing/decreasing of the function ƒ(x) between -1 and 2?

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We are being told that the instantaneous rate of change of the function is at least of 3 on the interval [-1,2], so the smallest possible value of ƒ(2) is the case where the function has the smallest rate of change, that is a constant rate of change of 3 which means it is a straight line (of slope 3). And since we are starting at the point (-1,9) with a slope of 3 and a run of 3 (from -1 to 2) we will end up with a rise of 9 and so the minimal value of ƒ(2) is 18.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Alternatively, you can use the MVT (Mean Value Theorem), which states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that $f'(c)={\frac {f(b)-f(a)}{b-a}}$ If we assume $f(x)$ is differentiable on (-1,2) and differentiable and continuous on the interval [-1,2], we can use the MVT which gives: ${\frac {f(2)-f(-1)}{2-(-1)}}=f'(c)$ For some value c in the interval (-1,2). However, we know that $f'(x)\geq 3$ for any value between -1 and 2, thus, $f'(c)\geq 3$. Adding this, plugging in known values, and simplifying gives: ${\frac {f(2)-9}{3}}\geq 3$ Solving for f(2) we get $f(2)\geq 18$, thus the minimal value of f(2) is 18.

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