Problem 4 - Calculus: Early Transcendentals

Question

Use \(f(x)=\sqrt[3]{x+1}\) to approximate \(\sqrt[3]{9}\) by choosing an appropriate point \(x=a\). Are we over- or under-estimating the value of \(\sqrt[3]{9}\) ? Explain.

Step-by-Step Solution

Answer

The approximation at \(a=8\) results in \(\frac{19}{9}\), which is an over-estimation of \(\sqrt[3]{9}\).

Step 1: Identify the Function and Target

The given function is \(f(x) = \sqrt[3]{x+1}\). We need to approximate \(\sqrt[3]{9}\). Notice that \(\sqrt[3]{9} = \sqrt[3]{(8+1)}\), so we should aim to expand around \(x = 8\) in order to approximate this value effectively.

Step 2: Choose a Point for Approximation

We want to use point \(a\) where \(a+1=9\), so we let \(a = 8\). Here, \(\sqrt[3]{8+1}\) or \(\sqrt[3]{9}\) can be easily approximated by using a linear approximation technique.

Step 3: Find the Derivative of the Function

To approximate \(f(x)\), compute its derivative. Given \(f(x) = \sqrt[3]{x+1}\), the derivative is \(f'(x) = \frac{1}{3}(x+1)^{-2/3}\).

Step 4: Evaluate the Function and its Derivative at the Chosen Point

Compute \(f(8)\) and \(f'(8)\). We have:\(f(8) = \sqrt[3]{9} = 2\)\(f'(8) = \frac{1}{3}(9)^{-2/3} = \frac{1}{3}\times \frac{1}{3} = \frac{1}{9}\).

Step 5: Formulate the Linear Approximation

Using the linear approximation formula, \(L(x) = f(a) + f'(a)(x - a)\), here, substituting \(a = 8\):\(L(x) = 2 + \frac{1}{9}(x - 8)\).

Step 6: Approximate \(\sqrt[3]{9}\) Using Linearization

Substitute \(x = 8\) in the linear approximation:\(L(8) = 2 + \frac{1}{9}(9 - 8) = 2 + \frac{1}{9} = \frac{19}{9}\).

Step 7: Determine if Over- or Under-Estimation

Calculate \(\sqrt[3]{9}\) using a calculator, which is approximately \(2.08008\). Comparing this to our linear approximation \(\frac{19}{9} = 2.111...\), the calculated linear approximation is slightly higher, thus it is an over-estimation.

Key Concepts

Linear Approximation

Linear approximation is a helpful tool in calculus that lets us estimate the value of a function around a point. It uses the idea of tangent lines to predict values of the function. Think of it like "zooming in" on a tiny piece of a curve until it looks straight.

It starts with a simple formula:

• The formula is \[L(x) = f(a) + f'(a)(x - a)\].
• Here, \(f(a)\) is the value of the function at the point of interest, \(a\), and \(f'(a)\) is the derivative at that same point.
• The (\(x - a\)) part helps tell us how far from the point \(a\) we want the approximation.

The formula suggests a linear path that can make complex calculations simpler. If we know the shape of our function at a particular spot, we can predict nearby values almost effortlessly. Linear approximation provides a path to simplicity in the world of complicated curves.

In the exercise, we saw how easy it was to estimate \(\sqrt[3]{9}\) using the linear approximation technique. By knowing \(f'(a)\), we refined a guess for a smooth curve into a factual point calculation.

Derivative of a Function

Derivatives are fascinating tools in calculus that tell us how things change. You can think of them as the speedometer of a function, showing how fast or slow it goes.

The derivative works by calculating the slope of a tangent line to a curve at any given point. This slope helps tell how a function grows or shrinks.

• Consider a simple tangent line as a rule for the curve's behavior at that spot.
• For any function \(f(x)\), the derivative \(f'(x)\) sends out an alarm when the function starts changing suddenly.
• In practice, to find \(f'(x)\), we may need to apply rules like the power rule, product rule, or chain rule.

In our exercise, we differentiated the cube root function to solve the linear approximation task. The derivative proved to be the key, \(f'(x) = \frac{1}{3}(x+1)^{-2/3}\), aiding in our calculation at \(x=8\). It told us the slope at that point, leading to an optimal approximation.

Cube Root Function

The cube root function, written as \(f(x) = \sqrt[3]{x}\), is intriguing. It solves the question, "What number times itself three times gives me this?"

• Let's break it down: \(\sqrt[3]{8} = 2\), because \(2 \times 2 \times 2 = 8\).
• Cube roots work on both positive and negative numbers, unlike square roots.
• Every real number has a real cube root.

In this exercise, we used the cube root function plus one: \(f(x) = \sqrt[3]{x+1}\). Choosing it was smart because the cube root function's simplicity lets us focus on accuracy when approximating nearby values. With linear approximation, we saw how the cube root at \(9\) smoothly estimates to just above \(2\), with the aid of calculus.
You could say the cube root function "plays nice" when fused with derivatives.