Science:Math Exam Resources/Courses/MATH102/December 2011/Question 01 (a) i

MATH102 December 2011

  •  Q1 (a) i  •  Q1 (a) ii  •  Q1 (b) i  •  Q1 (b) ii  •  Q1 (b) iii  •  Q1 (c)  •  Q2 (a) i  •  Q2 (a) ii  •  Q2 (a) iii  •  Q2 (b) i  •  Q2 (b) ii  •  Q2 (b) iii  •  Q2 (c)  •  Q3  •  Q4 (i)  •  Q4 (ii)  •  Q4 (iii)  •  Q4 (iv)  •  Q4 (v)  •  Q4 (vi)  •  Q5  •  Q6  •  Q7 (i)  •  Q7 (ii)  •  Q7 (iii)  •  Q8 (i)  •  Q8 (ii)  •  Q8 (iii)  •  Q9  •  

   Other MATH102 Exams

  •  December 2014  •  December 2011  •  December 2012  •  December 2013  •  December 2016  •  December 2015  •  

[hide] Question 01 (a) i
Short-Answer Questions. A correct answer in the box gives full marks. For partial marks work needs to be shown. The values of f(x) are as given in the table. x 1.00 1.50 2.00 2.50 3.00 f(x) 1.12 1.06 1.02 1.00 1.03 a. Based on the above table, provide the best estimate for f(x) and fill in the second table. x 1.25 1.75 2.25 2.75 f’(x) b. Estimate $\displaystyle f''(2.0)$

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!

[show] Hint
Remember that the derivative is the slope of the tangent line of a function at a point on the function. What would be a good way to estimate this line? Try drawing a picture to give you an idea.

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[show] Solution

Given the data we have, the best way to approximate the derivative is by using secant lines! For an example, let's first estimate $\displaystyle f'(1.25)$. For this, lets use the secant line from x=1 and x=1.5. This gives $\displaystyle f'(1.25)\approx {\frac {f(1.5)-f(1)}{1.5-1}}={\frac {1.06-1.12}{0.5}}={\frac {-0.06}{0.5}}=-0.12$. Proceeding as above, we can fill in the table as follows: x 1.25 1.75 2.25 2.75 f’(x) -0.12 -0.08 -0.04 0.06 For the second derivative, we use the same logic above except we do it with the derivative as our function. This gives $\displaystyle f''(2)\approx {\frac {f'(2.25)-f'(1.75)}{2.25-1.75}}={\frac {-0.04-(-0.08)}{0.5}}={\frac {0.04}{0.5}}=0.08$ and this completes the question.

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