Question

In Exercises \(1-6,\) (a) find the linearization \(L(x)\) of \(f(x)\) at \(x=a\) . (b) How accurate is the approximation \(L(a+0.1) \approx f(a+0.1) ?\) See the comparisons following Example \(1 .\) \(f(x)=x^{3}-2 x+3, \quad a=2\)

Short answer

The linearization \(L(x)\) of \(f(x) = x^3 - 2x + 3\) at \(x = 2\) is \(L(x) = 5 + 10(x-2)\). The approximation \(L(a+0.1)/f(a+0.1)\) is relatively accurate; the actual difference is only 0.061.

Step-by-step solution

Find the derivative of \(f(x)\)

The derivative of \(f(x) = x^3 - 2x + 3\) is \(f'(x) = 3x^2 - 2\).

Evaluate the derivative at \(x=a\)

Substitute \(a = 2\) into \(f'(x) = 3x^2 - 2\) to find \(f'(a) = 3(2)^2 - 2 = 10\).

Find the value of \(f(a)\)

Substitute \(a = 2\) into \(f(x) = x^3 - 2x + 3\) to find \(f(a) = (2)^3 - 2(2) + 3 = 5\).

Construct the linearization \(L(x)\)

From Steps 2 and 3, we have \(f'(a)=10\) and \(f(a)=5\). So the linearization \(L(x)\) at \(x=a\) is \(L(x) = f(a) + f'(a)(x-a) = 5 + 10(x - 2)\).

Find the approximation \(L(a+0.1)\)

Substitute \(a+0.1 = 2.1\) into \(L(x) = 5 + 10(x - 2)\) to find \(L(a + 0.1) = 5 + 10(2.1 - 2) = 6\).

Evaluate \(f(a+0.1)\)

Substitute \(a+0.1 = 2.1\) into \(f(x) = x^3 - 2x + 3\) to find \(f(a+0.1) = (2.1)^3 - 2(2.1) + 3 = 6.061\).

Compare \(L(a+0.1)\) with \(f(a+0.1)\)

Here, the approximation \(L(a+0.1) = 6\) and actual value \(f(a+0.1) = 6.061\). The absolute difference is \(|6.061-6| = 0.061\).

Key concepts

Derivative Calculation

The derivative of a function represents the rate at which the function's value changes as its input changes. This concept is crucial when you need to find the slope of the tangent line to the function's graph at a given point.
In our example, the derivative calculation starts with finding the derivative of the function, denoted as f'(x). For the function f(x)=x^3-2x+3, the derivative f'(x)=3x^2-2 is obtained by applying the power rule, which says to multiply the exponent by the coefficient and decrease the exponent by one for each term.

• The term x^3 becomes 3x^2.
• The term -2x translates to -2, since the derivative of x is 1.
• The constant 3 goes to 0, as the derivative of any constant is zero.

This step is the foundation for linearization and helps us establish the equation for the tangent line that will approximate our function around a specific point.

Tangent Line Approximation

Once the derivative at a point is calculated, we can proceed to find the equation of the tangent line at that point. The equation of the tangent line is the linear approximation or linearization of the function at a specific value of x=a. It is denoted by L(x).
For the function given by f(x) = x^3 - 2x + 3, the linearization at x=a=2 is calculated by evaluating the derivative at that point to get the slope, along with the function's value at that point to get the y-intercept. Specifically, L(x) = f(a) + f'(a)(x - a).
Thus, the equation L(x) = 5 + 10(x - 2) gives us a straight line that skims the curve of the function at x=2. This straight line can be used to approximate values of f(x) close to x=2, which is especially useful for making predictions or for calculations when the actual function is too complex.

Function Approximation Accuracy

The accuracy of the function approximation using the tangent line hinges on how close the x-value is to the point of tangency. The nearer the value is to a, the more accurate the approximation will be because it is based on local linearity – the idea that over a small interval, a curve can be closely approximated by a line.
In our exercise, we approximate f(a+0.1) using the linearization. The calculated value from the linearization, L(a+0.1) = 6, and the actual function value, f(a+0.1) = 6.061, show a small difference. The accuracy of the approximation is measured by the absolute error, which is |6.061 - 6| = 0.061 in this case.
This demonstrates that the tangent line approximation at a=2 reasonably estimates the function's value at x=2.1, with only a slight error. Therefore, accuracy is inherently linked to proximity to the point of tangency and to the function's curvature—the less curved the function is around the point of tangency, the more accurate the linear approximation will be.