Problem 32 - Calculus

Question

Values of \(x\) and \(g(x)\) are given in the table. For what value of \(x\) does \(g^{\prime}(x)\) appear to be closest to \(3 ?\) $$\begin{array}{c|c|c|c|c|c|c|c|c} \hline x & 2.7 & 3.2 & 3.7 & 4.2 & 4.7 & 5.2 & 5.7 & 6.2 \\ \hline g(x) & 3.4 & 4.4 & 5.0 & 5.4 & 6.0 & 7.4 & 9.0 & 11.0 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

\(x = 5.45\) is where \(g'(x)\) is closest to 3.

Step 1: Understand the Problem

We need to find a value of \(x\) where the derivative \(g'(x)\) is close to \(3\). From the table, we have discrete values of \(x\) and \(g(x)\). The derivative \(g'(x)\) represents the rate of change or slope of \(g(x)\) with respect to \(x\). We can approximate this by calculating the difference between Consecutive \(g(x)\) values divided by the difference in \(x\) values.

Step 2: Use the Slope Formula

For two consecutive points \((x_i, g(x_i))\) and \((x_{i+1}, g(x_{i + 1}))\), the slope \(m\) is approximated by:\[m = \frac{g(x_{i+1}) - g(x_i)}{x_{i+1} - x_i}\]We will apply this formula to each pair of consecutive points to find approximations of the derivatives \(g'(x)\).

Step 3: Calculate Derivatives for Each Interval

1. From \(x = 2.7\) to \(x = 3.2\): \[\frac{4.4 - 3.4}{3.2 - 2.7} = \frac{1}{0.5} = 2.0\]2. From \(x = 3.2\) to \(x = 3.7\): \[\frac{5.0 - 4.4}{3.7 - 3.2} = \frac{0.6}{0.5} = 1.2\]3. From \(x = 3.7\) to \(x = 4.2\): \[\frac{5.4 - 5.0}{4.2 - 3.7} = \frac{0.4}{0.5} = 0.8\]4. From \(x = 4.2\) to \(x = 4.7\): \[\frac{6.0 - 5.4}{4.7 - 4.2} = \frac{0.6}{0.5} = 1.2\]5. From \(x = 4.7\) to \(x = 5.2\): \[\frac{7.4 - 6.0}{5.2 - 4.7} = \frac{1.4}{0.5} = 2.8\]6. From \(x = 5.2\) to \(x = 5.7\): \[\frac{9.0 - 7.4}{5.7 - 5.2} = \frac{1.6}{0.5} = 3.2\]7. From \(x = 5.7\) to \(x = 6.2\): \[\frac{11.0 - 9.0}{6.2 - 5.7} = \frac{2.0}{0.5} = 4.0\]

Step 4: Identify the Closest Derivative to 3

Compare the calculated derivatives to determine which is closest to \(3\):- \(2.0\) at \(x = 2.95\)- \(1.2\) at \(x = 3.45\)- \(0.8\) at \(x = 3.95\)- \(1.2\) at \(x = 4.45\)- \(2.8\) at \(x = 4.95\)- \(3.2\) at \(x = 5.45\)- \(4.0\) at \(x = 5.95\)The slope \(3.2\) at \(x = 5.45\) is closest to \(3\).

Step 5: Conclude the Value of x

The value of \(x\) where \(g'(x)\) is closest to \(3\) is at the midpoint of the interval from \(x = 5.2\) to \(x = 5.7\), where the calculated derivative is \(3.2\). Thus, \(x = 5.45\).

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus used to find the derivative of a function, which represents the function's rate of change. It is a way to approximate the slope of the tangent line to the curve at any given point. The difference quotient is expressed as:

• \( \frac{g(x+h) - g(x)}{h} \)

Here, \(h\) is a small increment in \(x\). If \(h\) approaches zero, the difference quotient becomes the derivative of the function at that point.
This concept is essential in understanding how the function behaves locally and in determining how the function changes as \(x\) changes.
By using discrete values, as seen in the table, we can approximate this difference quotient between consecutive values, like from \(x = 2.7\) to \(x = 3.2\), to approximate the derivative at particular intervals.

Rate of Change

The rate of change indicates how one quantity changes in response to another quantity. In our context, it means how the function \(g(x)\) changes with respect to \(x\). Essentially, it's telling us how steep or how flat a curve is at a particular point. The concept is crucial when analyzing functions and their graphs.

• To calculate the rate of change, we use the slope calculation or the difference quotient.

Understanding the rate of change helps us make predictions about the behavior of the function in question. For example, if you see a high rate of change, the graph is steep, indicating rapid changes in the function's output. Conversely, a low rate of change suggests a flatter curve.
In solving exercises like the one given, recognizing areas of high or low rate of change helps us identify intervals where derivatives are close to a given value, like the slope of 3.

Slope Calculation

Slope calculation is the process of finding the steepness or incline of a line, and it involves determining the rate at which one variable changes in relation to another. The slope is crucial when discussing straight lines in coordinate geometry and serves as a fundamental concept when studying curves in calculus.
The slope between two points \((x_i, g(x_i))\) and \((x_{i+1}, g(x_{i+1}))\) on a graph can be calculated using the formula:

• \( m = \frac{g(x_{i+1}) - g(x_i)}{x_{i+1} - x_i} \)

This formula gives us an approximation of the derivative between those two points, helping us understand how the function is changing at specific intervals.
By applying this to consecutive points, we can systematically determine the slope of tangent lines to the curve represented by discrete data points, as was done step by step in our exercise. This method was used to find where the slope is closest to 3, specifically pinpointing an interval from \(x = 5.2\) to \(x = 5.7\), giving a slope of 3.2.