Problem 39 - Calculus

Question

Which points between \(t=0\) and \(t=5\) can be in the domain of \(f(t)\) ? With this domain find the range in \(37-42\). $$ f(t)=|t-4| \text { (absolute value) } $$

Step-by-Step Solution

Answer

No points in the range [37, 42] since range of \(f(t)\) is [0, 4].

Step 1: Understand the Function

The function is given by \( f(t) = |t-4| \). The absolute value function outputs the non-negative distance of \( t \) from 4. This means \( f(t) \) is always non-negative and the smallest value it can take is 0.

Step 2: Determine the Domain

The domain of \( f(t) = |t-4| \) is the set of all possible values of \( t \) for which the function is defined. Since \( f(t) \) is an absolute value function, it is defined for all real numbers. Therefore, the domain between \( t = 0 \) and \( t = 5 \) is all the values in that interval: \( 0 \leq t \leq 5 \).

Step 3: Evaluate the Function over the Domain

To find the range, we evaluate \( f(t) = |t-4| \) over the domain \( 0 \leq t \leq 5 \). Plugging in the boundary values and some interior values: \( f(0) = |0-4| = 4 \), \( f(4) = |4-4| = 0 \), and \( f(5) = |5-4| = 1 \). The function decreases from 4 to 0 as \( t \) moves from 0 to 4, then increases from 0 to 1 as \( t \) moves from 4 to 5.

Step 4: Determine the Possible Range

From the evaluated values, the range of the function \( f(t) \) when \( t \) is between 0 and 5 is \( [0, 4] \). Since we need to identify if the range intersects \([37, 42]\), there is no intersection because \([0, 4]\) does not overlap with \([37, 42]\). Thus, the function cannot reach any value between 37 and 42 in the specified domain.

Key Concepts

Absolute Value Function

The absolute value function is a fundamental concept in mathematics, often symbolized as \(\ |x|\ \). It measures the distance of a number from zero on the real number line.
This means it always returns a non-negative value. In other words, \(\ |x|\ \) equals \(x\) if \(x\) is positive or zero, but \(-x\) if \(x\) is negative.
For the function given in the exercise, \(f(t) = |t-4|\), it specifically measures how far any point \(t\) is from 4.

Key Points about Absolute Value Functions:

• Non-negative output: Always zero or positive.
• Distance representation: Reflects the distance from a specific point, in this case, point 4.
• Continuous and defined for all real numbers.

Understanding this property helps in identifying the range as the function will stretch from the point closest to zero outwards over its domain.

Real Numbers

Real numbers are the broad collection of numbers that we use in everyday math. They include:

• Natural numbers (1, 2, 3, ...)
• Whole numbers (0, 1, 2, 3, ...)
• Integers (..., -3, -2, -1, 0, 1, 2, 3, ...)
• Rational numbers (fractions like \(\ \frac{1}{2}, \frac{3}{4}\) )
• Irrational numbers (like \(\ \pi\) and \(\sqrt{2}\) )

Essentially, real numbers cover every point on the number line that you can think of.
Since \(f(t) = |t-4|\) involves real numbers, we consider all possible values of \(t\) within defined intervals like \(0\) to \(5\).
This inclusion means our function is defined at every single point along this continuum, making it easy to calculate and understand.

Interval Evaluation

Interval evaluation is a process of determining the function's behavior over a specified set of points.
In our problem, we look at the interval from \(t = 0\) to \(t = 5\).
This requires plugging various values of \(t\) from this interval into the function \(f(t) = |t-4|\).

Steps for Interval Evaluation:

• Identify the interval: here it is \(0 \leq t \leq 5\).
• Plug in boundary and key mid-point values into the function: \(f(0), f(4), f(5)\).
• Observe how the function values change as \(t\) varies.

In this case, we see the function decreases towards the minimum at \(t=4\), and then increases.
Understanding these changes helps in predicting the behavior of functions over different domains.

Function Evaluation

Function evaluation involves calculating outputs of a function based on given inputs.
With \(f(t) = |t-4|\), function evaluation helps us establish the domain's outputs and thus the function's range.
For each value of \(t\), substitute into \(f(t)\) to see the corresponding output.
For instance:

• Input: \(t = 0\), Output: \(f(0) = |0-4| = 4\)
• Input: \(t = 4\), Output: \(f(4) = |4-4| = 0\)
• Input: \(t = 5\), Output: \(f(5) = |5-4| = 1\)

This systematic substitution process creates a range of potential outputs, helping identify any intersections with other specified ranges.
It's a fundamental skill for predicting function characteristics and intersections with defined benchmarks, such as checking our range against \([37, 42]\).