Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Use the fact that \(\frac{d}{d x} x^{1 / 3}=\left(\frac{1}{3}\right) x^{-2 / 3}\) to approximate \(\sqrt[3]{30}\). Do you expect your answer to be an over- approximation or an under-approximation? Explain. Compare your answer to the approximation supplied by your calculator.

Step-by-Step Solution

Answer

From the detailed solution, the calculator will give a value higher than the one computed from the differential. Thus, it is an under-approximation.

Step 1: Identify the function, value, and derivative.

The function that is going to be worked upon is \(f(x) = x^{1/3}\). The value of \(x\) that is close to 30 which is a perfect cube is 27. The derivative at the point \(x_0 = 27\) to get the slope of the tangent line is \( f'(x_0) = (1/3)*27^{-2/3}\).

Step 2: Use linear approximation

Let's use the formula of linear approximation which is \(f(x) \approx f(x_0) + f'(x_0)(x - x_0)\). Substituting the values from Step 1, we have \(f(30) \approx 27^{1/3} + (1/3)*27^{-2/3}*(30-27)\). Now simplify this expression.

Step 3: Determine the type of approximation.

As the function \(x^{1/3}\) is increasing and concave, it means that a line tangent to the curve at \(x = 27\) underestimates the function at \(x = 30\). So, the answer we got in step 2 is an under-approximation.

Step 4: Compare the approximation to calculator evaluation.

Lastly, compute \(\sqrt[3]{30}\) using a calculator and compare it with the result obtained in step 2.

Key Concepts

Derivative

The concept of a derivative is central to calculus and helps us understand how a function behaves. It's like figuring out how fast something is changing at any point in time. When we talk about the derivative of a function at a certain point, we're referring to the slope of the tangent line at that point on the graph.
For the exercise at hand, we have a function noted as \(f(x) = x^{1/3}\), which is the cube root function. The derivative of this function, \(f'(x)\), represents how steep the slope of the tangent line to \(f(x)\) is, or in simpler terms, how quickly \(f(x)\) is increasing or decreasing at any point.
In the given problem, we calculate this derivative as \(f'(x) = \left(\frac{1}{3}\right) x^{-2/3}\). This tells us about the rate of change or how the cube root of \(x\) changes as \(x\) varies.

Tangent Line

The tangent line is a straight line that just "touches" a curve at a given point, without crossing it. Imagine this line as a best approximation of the curve right at the point of tangency. It is very useful for approximating the values of a function near that point.
In our linear approximation exercise, we start by finding a point on the puzzle piece. For \(x = 27\), we identify this on the curve of \(f(x) = x^{1/3}\). The slope of the tangent line at \(x = 27\) is given by the derivative \(f'(27)\).
Using this tangent, we approximate \(\sqrt[3]{30}\) by estimating how the cube root function changes from \(x = 27\) to \(x = 30\). The tangent provides a linear approximation through the formula:

• \(f(x) \approx f(x_0) + f'(x_0)(x - x_0)\)

where \(x_0 = 27\). This formula finds a close approximation to the actual function value, making complex functions simpler to work with.

Concave Functions

Concavity describes the "bending" of a curve. A function that is concave up looks like a cup and holds water, while a function that is concave down looks like an upside-down cup. Concave functions have important roles in various mathematical analyses.
For the function \(f(x) = x^{1/3}\), the function exhibits concavity which affects how well a tangent line can approximate the function. In this case, the function is concave on the interval we're concerned with, which causes the tangent line to sit entirely below the curve.
This means that the straight-line approximation we made using the tangent line will usually be less than the actual function value at \(x = 30\). Thus, indicating that our approximation is an under-estimation of the true cube root value. Recognizing the concavity helps in understanding the accuracy of linear approximations.

Cube Roots

Cube roots can initially appear complex, but they are a way of returning to the original "cubed" number. Specifically, the cube root of a number \(x\) is a number \(y\) such that \(y^3 = x\). It’s the value that, when multiplied by itself twice, gives \(x\).
For instance, \(\sqrt[3]{27} = 3\), because \(3 \times 3 \times 3 = 27\). In this exercise, we're using this principle to estimate \(\sqrt[3]{30}\) since 30 is not a perfect cube. By using linear approximation, we leverage the nearby known value of \(\sqrt[3]{27}\).
By approximating, we're seeking ease, as direct complex calculation might be difficult without a calculator. Hence, the reliability of cube root approximations depends greatly on adjacent perfect cubes and understanding the behavior of the cube root function near these points.