Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (b)

MATH104 December 2016

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Question 04 (b)
Decide whether the estimate from 4(a) is an over or under estimate and then use the formula, ${\frac {M}{2}}\left|x-a\right|^{2}$, to determine the worst case error.

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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 approximation underestimates for a concave up curve, and overestimates for one that is concave down. As for error, use the formula $|{\mbox{Error}}|\leq {\frac {M}{2}}(x-a)^{2},$ where $M>0$, to estimate an error bound for your approximation. To find a value for $M$, it would suffice to find an upper bound on the absolute value of the second derivative and use this as the value for $M$.

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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. The above is an over-estimate since the graph of $y=f(x)={\sqrt {x}}$ is concave down. Then to find $M$ we need to maximize $g(x)=\left|f''(x)\right|=\left|{\frac {-1}{4}}x^{-1.5}\right|={\frac {1}{4x^{1.5}}}$ on the interval $\left[64,69\right]$. Since the denominator increases as $x$ increases, the fraction will be maximized with the smallest denominator, i.e. at the left endpoint. Alternatively, $g'(x)={\frac {1}{4}}\cdot {\frac {-3}{2}}x^{-2.5}<0$. Therefore, $g(x)$ is always decreasing, which makes $M={\frac {1}{4\cdot (64)^{1.5}}}={\frac {1}{4\cdot 8^{3}}}$, and the worst-case error to be $\left|E\right|\leq {\frac {M}{2}}(x-a)^{2}={\frac {1}{8^{4}}}\cdot 5^{2}$ answer: $\color {blue}{\frac {1}{8^{4}}}\cdot 5^{2}$