Question:

A function f(x) is defined as $f(x)=\sqrt[3]{x^2+1}$.

(a) Find the linear approximation of f(x) at the point x = 2.

(b) Use the linear approximation to estimate the value of f(2.1) and calculate the maximum error in this estimation.

Answer:

(a) To find the linear approximation of f(x) at the point x = 2, we will begin by finding the equation of the tangent line at x = 2.

1. Find the derivative of f(x): $f'(x) = \frac{d}{dx}\left(\sqrt[3]{x^2+1}\right)$

Using the chain rule, we have:

$f'(x) = \frac{1}{3}(x^2+1)^{-2/3} \cdot 2x$

Simplifying further:

$f'(x) = \frac{2x}{3(x^2+1)^{2/3}}$

2. Evaluate the derivative at x = 2:

$f'(2) = \frac{2 \cdot 2}{3(2^2+1)^{2/3}}$

$f'(2) = \frac{4}{3(5)^{2/3}}$

3. Find the equation of the tangent line using the point-slope form.

The equation of the tangent line at x = 2 will be in the form:

$y = f(2) + f'(2)(x-2)$

We need to find f(2) to complete our equation.

$f(2) = \sqrt[3]{2^2+1}$

$f(2) = \sqrt[3]{5}$

So, the equation of the tangent line is:

$y = \sqrt[3]{5} + \frac{4}{3(5)^{2/3}}(x-2)$

(b) Now, we will use the linear approximation equation we derived in part (a) to estimate the value of f(2.1).

Substituting x = 2.1 into the equation of the tangent line, we have:

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3(5)^{2/3}}(2.1-2)$

Simplifying:

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3(5)^{2/3}}(0.1)$

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3} \cdot \frac{0.1}{(5)^{2/3}}$

Calculating the approximation:

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3} \cdot \frac{0.1}{\sqrt[3]{5^2}}$

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3} \cdot \frac{0.1}{\sqrt[3]{25}}$

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3} \cdot \frac{0.1}{\sqrt[3]{5} \cdot \sqrt[3]{5}}$

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3} \cdot \frac{0.1}{\sqrt[3]{5} \cdot 5^{2/3}}$

Calculating the value:

$f(2.1) \approx \sqrt[3]{5} + \frac{4}{3} \cdot \frac{0.1}{\sqrt[3]{5} \cdot 5^{2/3}} \approx 1.8172$

To calculate the maximum error in this estimation, we can use the formula for the error of a linear approximation:

$Error = |f(x) - L(x)|$

where f(x) is the original function and L(x) is the linear approximation.

For f(x) = $\sqrt[3]{x^2+1}$ and L(x) = $\sqrt[3]{5} + \frac{4}{3(5)^{2/3}}(x-2)$, we can find the maximum error by calculating:

$Error = \left| f(2.1) - L(2.1)\right|$

Substituting the values, we have:

$Error = \left| \sqrt[3]{2.1^2+1} - \left(\sqrt[3]{5} + \frac{4}{3(5)^{2/3}}(2.1-2)\right)\right|$

$Error = \left| \sqrt[3]{4.41+1} - \left(\sqrt[3]{5} + \frac{4}{3(5)^{2/3}}(0.1)\right)\right|$

$Error = \left| \sqrt[3]{5.41} - \left(\sqrt[3]{5} + \frac{4}{3(5)^{2/3}}(0.1)\right)\right|$

$Error \approx | 1.8436 - 1.8172| \approx 0.0264$

Therefore, the estimated value of f(2.1) using the linear approximation is approximately 1.8172 with a maximum error of approximately 0.0264.