Science:Math Exam Resources/Courses/MATH110/April 2018/Question 08 (a)

MATH110 April 2018

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Question 08 (a)
In this question you will approximate the value of ${\textstyle e^{0.5}}$ in two different ways and answer a few questions about your calculations. Make sure you justify your answers. Answers without justification will not receive any points. You may assume that ${\textstyle e=2.718281...}$ . (a) Approximate ${\textstyle e^{0.5}}$ using a linear approximation of ${\textstyle f(x)=e^{x}}$ at ${\textstyle a=0}$ . Simplify your answer.

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
Recall that the formula for linear approximation is given by ${\textstyle f(x)\approx f(a)+f'(a)(x-a)}$ .

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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. We use the formula ${\textstyle f(x)\approx f(a)+f'(a)(x-a)}$ , as given in the hint. Using ${\textstyle a=0}$ , we have ${\textstyle f(a)=e^{0}=1}$ and ${\textstyle f'(a)=f(a)=1}$ . Plugging these into the formula gives $f(x)\approx 1+1(0.5-0)=1+0.5=1.5.$ Answer: The linear approximation gives ${\textstyle {\color {blue}e^{0.5}\approx 1.5}}$ .