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In this case $f(x) =\\;$ [[0]].

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$a= \\;$[[1]] and $h=\\; $[[2]]
Input your estimation to $5$ decimal places: [[3]]

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True value is: [[4]] (input to 5 decimal places).

\n ", "showCorrectAnswer": true, "marks": 0}], "statement": "

Use the approximation $f(a+h) \\approx f(a)+hf^{\\prime}(a)$ to estimate \\[\\var{a+h}^{\\frac{\\var{m}}{\\var{n}}} \\]for a suitable function $f(x)$.

", "tags": ["application of differentiation", "approximation", "approximation of the value of a function using the tangent", "approximations", "calculus", "Calculus", "checked2015", "equation of tangent", "first order approximation", "functions", "maclaurin series", "MacLaurin series", "mas1601", "MAS1601", "tangent equation"], "rulesets": {"std": ["all", "!collectNumbers"], "dpoly": ["std", "fractionNumbers"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \t\t

3/07/2012

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Added tags

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20/06/2012:

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Added tags.

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Got rid of request for 5dps for the function!

\n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "

Approximate $f(x)=(a+h)^{m/n}$ by $f(a)+hf^{\\prime}(a)$ to 5 decimal places and compare with true value.

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We have $f(x)=x^\\frac{\\var{m}}{\\var{n}}$ and $a= \\var{a}$, $h=\\var{h}$.
Note that $\\var{a}^\\frac{1}{\\var{n}}=\\var{c}$ and so using the approximation :
$f(a+h)\\approx f(a)+hf^{\\prime}(a)$ and $f^{\\prime}(x) = \\frac{\\var{m}}{\\var{n}}\\simplify[std]{x^({m-n}/{n})}$
we have:
\\[\\simplify[std]{{a+h}^({m}/{n})}\\approx \\simplify[simplifyFractions]{{a}^({m}/{n})+{h}*({m}/{n})*{a}^({p}/{n})}=\\simplify[std,!sqrtSquare]{{c}^{m}+ {h}*({m}/{n})*{c}^{m-n}}=\\var{est}\\]
to 5 decimal places.
The true value to 5 decimal places is {tr}.

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