Problem 19 - The Calculus with Analytic Geometry

Question

Find the absolute extrema of the given function on the given interval, if there are any, and find the values of \(x\) at which the absolute extrema occur. Draw a sketch of the graph of the function on the interval. $$ f(x)=|x-4|+1 ;(0,6) $$

Step-by-Step Solution

Answer

The absolute maximum value is 5 at \( x = 0 \). The absolute minimum value is 1 at \( x = 4 \).

Step 1 - Understanding the Absolute Value Function

Recognize that the function given is an absolute value function. The function can be expressed as two separate linear functions: 1. For \( x \geq 4\), \( f(x) = (x - 4) + 1 = x - 3 \) 2. For \( x < 4\), \( f(x) = -(x - 4) + 1 = 5 - x \). Note the point of change at \( x = 4 \) which is critical to finding extrema.

Step 2 - Identify Critical Points

Find the critical points within the interval (0, 6). These will include the endpoints of the interval and the point where the function changes its form. Thus, check \( x = 0 \), \( x = 4 \), and \( x = 6 \).

Step 3 - Evaluate the Function at Critical Points

Evaluate \( f(x) \) at each critical point: 1. \( f(0) = |0 - 4| + 1 = 4 + 1 = 5\) 2. \( f(4) = |4 - 4| + 1 = 0 + 1 = 1\) 3. \( f(6) = |6 - 4| + 1 = 2 + 1 = 3\)

Step 4 - Determine the Absolute Extrema

Compare the function values obtained at the critical points: \( f(0) = 5 \), \( f(4) = 1 \), and \( f(6) = 3 \). The absolute maximum value is \ 5 \ at \( x = 0 \), and the absolute minimum value is \ 1 \ at \( x = 4 \).

Step 5 - Sketch the Graph

Draw the graph using the evaluated points and the linear pieces of the function. - For \( x < 4 \), draw the line segment from \( (0, 5) \) to \( (4, 1) \). - For \( x \geq 4 \), draw the line segment from \( (4, 1) \) to \( (6, 3) \).

Key Concepts

absolute value function analysis

The given function is \( f(x) = |x-4| + 1 \). Recognizing it as an absolute value function is key. This class of functions has a distinctive 'V' shape when graphed. Let's break down \( f(x) \) into two linear expressions based on the critical point where the value inside the absolute value changes sign.
For \( x \geq 4 \), the function simplifies to \( f(x) = (x-4) + 1 = x - 3 \). For \( x < 4 \), the function turns into \( f(x) = -(x-4) + 1 = 5 - x \). Thus, for a given interval, the absolute value function can be expressed as two different linear functions, changing forms at an important point. Here, \( x = 4 \) is our boundary of interest, as it alters the function's form. Essentially, analyzing an absolute value function requires splitting it into linear pieces based on the point where the inner expression equals zero, making further calculations straightforward.

critical points evaluation

To find the extrema, we need to evaluate the function at specific points within the given interval \( (0, 6) \). These points include the interval's endpoints and any points where the function changes its expression. In our case, these are \( x = 0 \), \( x = 4 \) (where the function's form changes), and \( x = 6 \).
Consider \( f(x) \) at each critical point:

• At \( x = 0 \), \( f(0) = |0 - 4| + 1 = 5 \)
• At \( x = 4 \), \( f(4) = |4 - 4| + 1 = 1 \)
• At \( x = 6 \), \( f(6) = |6 - 4| + 1 = 3 \)

Comparing these values helps us identify the absolute maximum and minimum values: the maximum is \( 5 \) at \( x = 0 \), and the minimum is \( 1 \) at \( x = 4 \).

graph sketching in calculus

Sketching the function \( f(x) = |x-4| + 1 \) involves plotting points and drawing lines based on the critical points and function expressions found previously.
For \( x \geq 4 \), plot from \( (4,1) \) to \( (6,3) \) using the line segment \( y = x - 3 \). For \( x < 4 \), draw the line segment from \( (0,5) \) to \( (4,1) \) as dictated by the function \( y = 5 - x \). These segments fit together to create the typical 'V' shape of the absolute value function. This step visually demonstrates how the function behaves across the interval and verifies the extrema points previously calculated.
Remember to plot carefully, showing all crucial points derived during the analysis.