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sashaice [31]
3 years ago
11

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 ≈
Mathematics
1 answer:
Lisa [10]3 years ago
7 0

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

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