Answer:
[tex]L(x)=1+\dfrac{1}{3}x[/tex]
[tex]\sqrt[3]{0.95} \approx 0.9833[/tex]
[tex]\sqrt[3]{1.1} \approx 1.0333[/tex]
Step-by-step explanation:
Given the function: [tex]g(x)=\sqrt[3]{1+x}[/tex]
We are to determine the linear approximation of the function g(x) at a = 0.
Linear Approximating Polynomial,[tex]L(x)=f(a)+f'(a)(x-a)[/tex]
a=0
[tex]g(0)=\sqrt[3]{1+0}=1[/tex]
[tex]g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}[/tex]
Therefore:
[tex]L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x[/tex]
(b)[tex]\sqrt[3]{0.95}= \sqrt[3]{1-0.05}[/tex]
When x = - 0.05
[tex]L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833[/tex]
[tex]\sqrt[3]{0.95} \approx 0.9833[/tex]
(c)
(b)[tex]\sqrt[3]{1.1}= \sqrt[3]{1+0.1}[/tex]
When x = 0.1
[tex]L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333[/tex]
[tex]\sqrt[3]{1.1} \approx 1.0333[/tex]
