Science:Math Exam Resources/Courses/MATH110/April 2019/Question 07 (a)

MATH110 April 2019

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Question 07 (a)
Using a linear approximation of $f(x)=x^{1/3}$ at $a=8$, estimate $(7)^{1/3}$. 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
Given a differentiable function $f(x)$, the linear approximation of $f(x)$ at the point $x=a$ is a function $L(x)=kx+m$ which passes through the point $(a,f(a))$ and has the same slope as $f$ at that point. Knowing this information, apply it to the function $f(x)=x^{1/3}$ with $a=8$.

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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 are asked to use the linear approximation of $f(x)=x^{1/3}$ at the point $a=8$ to approximate $(7)^{1/3}$. This linear approximation, which we denote by $L(x)$, is given by $L(x)=f(a)+f'(a)(x-a)$. We already know that $a=8,f(x)=x^{1/3}$, so we need only find $f'(x)$. To do so, we apply the power rule, and find that $f'(x)={\frac {d}{dx}}x^{1/3}={\frac {1}{3}}x^{-2/3}$. Substituting this information into our formula for $L(x)$, we have $L(x)=(8)^{1/3}+{\frac {1}{3}}(8)^{-2/3}(x-8)=2+{\frac {1}{12}}(x-8)$. Then, to estimate the value of $7^{1/3}$, we plug in $x=7$ into our linear approximation. This gives $(7)^{1/3}\approx L(7)=2+{\frac {1}{12}}(7-8)={\frac {23}{12}}$.