Science:Math Exam Resources/Courses/MATH102/December 2012/Question C 03 (b)

MATH102 December 2012

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Question C 03 (b)
This question is about approximating the value of $b={\sqrt {48}}$. (b) Use a linear approximation to estimate b. Is your estimate larger or smaller than the actual value? Justify your claim.

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
The linear approximation formula is $f(x)\approx f'(a)(x-a)+f(a)$, where $a$ is a number near the number you're approximating. Based on part (a) of this question, what would be a good choice for 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. Using the linear approximation formula $f(x)\approx f'(a)(x-a)+f(a)$ where our function is $f(x)={\sqrt {x}}$ and our value of a will be a number near 48. Choosing a = 49 is sensible because you can take the square root easily. We can either calculate $\displaystyle f'(x)$ again or note that we already found it in question C2 as $f'(x)=1/(2{\sqrt {x}})$. Plugging all this in gives: $f(x)\approx {\frac {1}{2{\sqrt {49}}}}(x-49)+{\sqrt {49}}={\frac {1}{14}}(x-49)+7$ To approximate $b={\sqrt {48}}$, we say ${\sqrt {48}}=f(48)\approx {\frac {1}{14}}(48-49)+7={\frac {-1}{14}}+7=6{\frac {13}{14}}$ Whether the linear approximation is an over or under estimate depends on the concavity of the function - if $f(x)$ is concave up, the tangent line approximation lies below the graph and is an under-estimate; the opposite is true if $f(x)$ is concave down. Because $f(x)={\sqrt {x}}$ is concave down, our estimate in this problem is an over-estimate.

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