Question

Let \(x=1\) and \(\Delta x=0.01\). Find \(\Delta y\). \(f(x)=\frac{4}{\sqrt[3]{x}}\)

Short answer

\(\Delta y\) is given by \(\frac{4}{\sqrt[3]{1.01}} - 4\).

Step-by-step solution

Calculate \(f(x)\) and \(f(x + \Delta x)\)

Firstly, compute the function values \(f(x)\) and \(f(x + \Delta x)\). Here, \(x=1\) and \(\Delta x=0.01\). So, plug these values into the function \(f(x) = \frac{4}{\sqrt[3]{x}}\). We get1. \(f(x) = \frac{4}{\sqrt[3]{1}} = 4\)2. \(f(x + \Delta x) = \frac{4}{\sqrt[3]{1.01}}\)

Calculate \(\Delta y\)

Now compute \(\Delta y\) by applying the formula \(\Delta y = f(x + \Delta x) - f(x)\). As calculated in step 1,\(\Delta y = \frac{4}{\sqrt[3]{1.01}} - 4\)

Key concepts

Function Evaluation

Function evaluation refers to the process of computing the output of a given function for specific input values. In this exercise, we have the function \( f(x) = \frac{4}{\sqrt[3]{x}} \). At \( x = 1 \), we evaluate the function to find \( f(1) \), which simplifies to:

• Since \( \sqrt[3]{1} = 1 \), the expression becomes \( \frac{4}{1} = 4 \).

Function evaluation is fundamental in calculus as it provides specific values that help in understanding the behavior of functions over different ranges. Here, calculating the next function value \( f(x + \Delta x) \), where \( \Delta x = 0.01 \) and \( x = 1 \), requires plugging these into the function:

• Calculate \( \sqrt[3]{1.01} \), then divide 4 by this result to get \( f(1.01) \).

These steps demonstrate how small changes in \( x \) can affect the output, which is crucial when analyzing functions and their responsiveness to input changes.

Change in Variables

Change in variables involves understanding how a small change in the input, such as \( \Delta x \), influences the change in the output, represented as \( \Delta y \). This concept is central in differential calculus for analyzing how functions react to small variations. Use this method to approximate how the function behaves as its input changes slightly from its original value.In our problem:

• \( x = 1 \) changes to \( x + \Delta x = 1.01 \).
• Calculate \( f(x+\Delta x) - f(x) \) to determine \( \Delta y \), the change in the function's value, illustrating the change in variables concept.
Understanding these shifts allows students to grasp how the derivative, a principle in calculus, is determined. It essentially measures slopes of tangent lines at once value of \( x \), helping to predict behavior in nearby segments.

Calculus Problems

Calculus problems often involve determining how functions change and are evaluated at specific points. Calculus, especially differential calculus, focuses on rates of change which includes examining how one quantity changes concerning another. In typical calculus problems, such as the provided exercise, you observe how slight modifications in \( x \) (the independent variable) affect \( y \) (the dependent variable).For the function \( f(x) = \frac{4}{\sqrt[3]{x}} \) here:

• Given \( \Delta x = 0.01 \), find \( \Delta y \) using \( \Delta y = f(x+ \Delta x) - f(x) \).

This introduces learners to thinking critically about what these solutions mean in broader terms. Calculus problems teach the skills needed to infer insights about the nature of changes leading to advanced topics like integration, which considers accumulating those changes over an interval. This foundational understanding helps tackle real-world problems that involve dynamic change across different fields, such as physics or engineering.