Science:Math Exam Resources/Courses/MATH110/April 2019/Question 07 (a)/Solution 1

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 $(7)^{1/3}\approx L(7)=2+{\frac {1}{12}}(7-8)={\frac {23}{12}}$.