Problem 80 - Calculus Single Variable

Question

You are given a function \(f,\) a value \(c\) and \(a\) viewing rectangle \(R\) containing the point \(P=(c, f(c))\). In \(R,\) graph the four functions \(f, g, h,\) and \(k,\) where \(g(x)=\) \(f(c)+f^{\prime}(c)(x-c), h(x)=g(x)+1 / 2 f^{\prime \prime}(c)(x-c)^{2},\) and \(k(x)=\) \(h(x)+1 / 6 f^{\prime \prime \prime}(c) \quad(x-c)^{3} .\) The graphs of \(g, h,\) and \(k\) are, respectively, linear, parabolic, and cubic approximations of the graph of \(f\) near \(c\). The method of constructing these approximating functions, which are called Taylor polynomials of \(f\) with base point \(c,\) is studied in Chapter 8. \(f(x)=x^{3 / 2}, c=1, R=[0,4] \times[-1 / 2,9]\)

Step-by-Step Solution

Answer

Graph the functions \(f(x) = x^{3/2}\), \(g(x)\), \(h(x)\), and \(k(x)\) to see their approximations on the specified rectangle.

Step 1: Define Functions

To graph the functions, first define the functions given the parameters. We have:- Function \(f(x) = x^{3/2}\).- Linear approximation \(g(x) = f(c) + f'(c)(x-c)\).- Quadratic approximation \(h(x) = g(x) + \frac{1}{2} f''(c)(x-c)^2\).- Cubic approximation \(k(x) = h(x) + \frac{1}{6} f'''(c)(x-c)^3\).

Step 2: Function Evaluation at c

Evaluate the function \(f\) and its derivatives at the point \(c = 1\):- \(f(c) = 1^{3/2} = 1\)Calculate the derivatives:- \(f'(x) = \frac{3}{2}x^{1/2}\) so \(f'(c) = \frac{3}{2}(1)^{1/2} = \frac{3}{2}\).- \(f''(x) = \frac{3}{4}x^{-1/2}\) so \(f''(c) = \frac{3}{4}(1)^{-1/2} = \frac{3}{4}\).- \(f'''(x) = -\frac{3}{8}x^{-3/2}\) so \(f'''(c) = -\frac{3}{8}(1)^{-3/2} = -\frac{3}{8}\).

Step 3: Construct Taylor Polynomials

Using the evaluated values, construct the Taylor polynomials:- Linear: \(g(x) = 1 + \frac{3}{2}(x-1)\).- Quadratic: \(h(x) = g(x) + \frac{1}{2} \cdot \frac{3}{4} (x-1)^2 = 1 + \frac{3}{2}(x-1) + \frac{3}{8}(x-1)^2\).- Cubic: \(k(x) = h(x) + \frac{1}{6}(-\frac{3}{8})(x-1)^3 = 1 + \frac{3}{2}(x-1) + \frac{3}{8}(x-1)^2 - \frac{1}{16}(x-1)^3\).

Step 4: Graph the Functions

To visualize, graph the following functions on the viewing rectangle \([0,4] \times [-1/2,9]\):1. \(f(x) = x^{3/2}\).2. \(g(x) = 1 + \frac{3}{2}(x-1)\) (a linear approximation).3. \(h(x) = 1 + \frac{3}{2}(x-1) + \frac{3}{8}(x-1)^2\) (a parabolic approximation).4. \(k(x) = 1 + \frac{3}{2}(x-1) + \frac{3}{8}(x-1)^2 - \frac{1}{16}(x-1)^3\) (a cubic approximation). Each function provides a closer approximation to \(f(x)\) near \(c = 1\).

Key Concepts

Linear Approximation

In calculus, linear approximation is a method of estimating the value of a function near a specific point using only the first derivative. It's like finding a tangent line that just lightly touches the curve at that point.

Imagine you have a curvy path and that you want a straight surface to roll your toy car perfectly at one specific point, this straight surface would be your linear approximation. It simplifies the curve to a straight line, making calculations much easier.

The formula for linear approximation states that for a function \(f(x)\) and a point \(c\), the linear approximation \(g(x)\) is given by:

• \(g(x) = f(c) + f'(c)(x-c)\)

Here, \(f'(c)\) is the derivative of the function at point \(c\), showing us how the function grows or shrinks at that exact point.

This method is often used in situations where a quick estimation is required, or when the changes between values are small.

Quadratic Approximation

Quadratic approximation steps up from linear approximation by also considering the second derivative of the function. If linear approximation is a straight line, quadratic approximation is a simple, yet smooth curve, resembling a parabola.

Using quadratic approximation gives a more accurate estimation, especially for functions where the curve isn't very steep. This is because it captures more information about how the function behaves near the point of interest.

The formula for the quadratic approximation of a function \(f(x)\) at point \(c\) is:

• \(h(x) = f(c) + f'(c)(x-c) + \frac{1}{2}f''(c)(x-c)^2\)

Here, \(f''(c)\) is the second derivative, and it helps to include the curvature of the function in our estimation.

Just like linear approximation, quadratic approximation can be very helpful in approximating functions near a point but provides more precision when the changes of the function become significant.

Cubic Approximation

By entering the world of cubic approximation, you get even closer to the function's actual shape. Unlike linear or quadratic approximation, cubic approximation uses the third derivative to fine-tune how we approximate the function.

The cubic function allows not only direction and curvature, but also changes in the rate of curvature, making it a powerful tool for particularly wiggly functions.

The formula for a cubic approximation of \(f(x)\) at point \(c\) is:

• \(k(x) = f(c) + f'(c)(x-c) + \frac{1}{2}f''(c)(x-c)^2 + \frac{1}{6}f'''(c)(x-c)^3\)

In this formula, \(f'''(c)\) gives further refinement by accounting for changes in the curve's bending, making predictions even more accurate within the vicinity of \(c\).

Cubic approximations are most useful when the function exhibits significant variation over a small interval, providing a detailed representation of the function's behavior.

Function Derivatives

Derivatives are the fundamental building blocks in calculus for understanding how functions change. The derivative at a point gives the slope of the tangent line to the function's graph at that point.

When we calculate derivatives, we're essentially asking: "How does this function curve?" Through derivatives, we understand the speed, angle, and direction at which the function's value changes.

Consider the derivatives in the context of approximations:

• First Derivative \(f'(x)\): Tells us the rate of change or slope of the function; essential for linear approximation.
• Second Derivative \(f''(x)\): Reveals the function's concavity and is used in quadratic approximation to show curvature.
• Third Derivative \(f'''(x)\): Provides information about the change in concavity, refining our cubic approximation.

Understanding derivatives is critical when studying Taylor polynomials, as they're used in progressively more advanced approximations to capture the behavior of functions more precisely.