Question

Find the linear approximation of the function g(x) = 3 1 + x at a = 0. g(x) ≈ Use it to approximate the numbers 3 0.95 and 3 1.1 . (Round your answers to three decimal places.) 3 0.95 ≈ 3 1.1 ≈

Answer 1

Answer:

$\sqrt[3]{0.95} \approx 0.983\\\\\\\sqrt[3]{1.1} \approx 1.033$

Step-by-step explanation:

The function is:

$g(x) =\sqrt[3]{1+x}$

We can find a linear aproximation around a point with a Taylor's series:

$g(x)\approx g(a)+g'(a)(x-a)$

The value of g(a) is

$g(a)=g(0)=\sqrt[3]{1+0}=1$

We have to calculate the firste derivative of g(x):

$g'(x)=\frac{d}{dx}[(1+x)^ {1/3}]=(1/3)(1+x)^{-2/3}=\frac{1}{3(1+x)^{2/3}}$

Then, we calculate g'(a)

$g'(a)=g'(0)=\frac{1}{3(1+0)^{2/3}}=\frac{1}{3*1}= \frac{1}{3}$

The linear approximation is:

$g(x)\approx1+\frac{1}{3} (x-0)=\frac{x}{3}$

Calculate $\sqrt[3]{0.95}$

We can use the linear approximation to calculate this, with x=-0.05.

$0.95=1+x\rightarrow x=-0.05\\\\\sqrt[3]{0.95} \approx 1+\frac{(-0.05)}{3} =1-0.017=0.983$

Calculate $\sqrt[3]{1.1}$

We can use the linear approximation to calculate this, with x=0.1.

$1.1=1+x\rightarrow x=0.1\\\\\sqrt[3]{1.1} \approx 1+\frac{(0.1)}{3} =1+0.033=1.033$