Problem 37 - 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 \(37-42\). $$ f(t)=\sqrt{t-1} $$

Step-by-Step Solution

Answer

The function's range in the given domain is [0, 2], none of which lies within [37, 42].

Step 1: Understand the Function

The function given is \(f(t) = \sqrt{t-1}\). We need to ensure that the expression under the square root sign is non-negative, i.e., \(t-1 \geq 0\).

Step 2: Solve the Inequality

To find which values of \(t\) can be used in the function, we solve the inequality \(t-1 \geq 0\). This gives us \(t \geq 1\).

Step 3: Apply Time Interval Constraint

Since we are asked to only consider the time interval between \(t=0\) and \(t=5\), we combine this with our inequality. Thus, \(1 \leq t \leq 5\). This is the domain of \(f(t)\) within the given interval.

Step 4: Determine the Range

For \(f(t) = \sqrt{t-1}\), using the domain \([1, 5]\), calculate the minimum and maximum values: \(\min(f(t)) = \sqrt{1-1} = 0\) and \(\max(f(t)) = \sqrt{5-1} = 2\). Therefore, the range of \(f(t)\) is \([0, 2]\).

Step 5: Check if the Range is within [37, 42]

The obtained range of the function is \([0,2]\). It does not overlap with the interval \([37, 42]\), so no values from the range are within this specified range.

Key Concepts

Understanding Inequalities

In mathematics, inequalities are statements about the relative size or order of two objects. They use symbols such as \(\geq\) (greater than or equal to) and \(\leq\) (less than or equal to).

These are critical when working with functions like \(f(t) = \sqrt{t-1}\), where we need to ensure that the expression under the square root is non-negative.

• This condition ensures that the square root is defined for real numbers.
• For \(t-1\) when \(f(t) = \sqrt{t-1}\), it means \(t-1 \geq 0\).

Understanding how inequalities affect the domain of a function is essential because it tells you which values of \(t\) are permissible for the function to output real values. Solving \(t-1 \geq 0\) yields \(t \geq 1\), which establishes part of the domain for the function.

Finally, combining this result with a given interval like \(0 \leq t \leq 5\) ensures that all mathematical conditions meet real-world constraints, or any added restrictions.

Function Analysis of \(f(t) = \sqrt{t-1}\)

Function analysis involves understanding the behavior and properties of mathematical functions. For \(f(t) = \sqrt{t-1}\), the analysis focuses on identifying its domain and range, which is crucial for grasping how the function operates across different intervals.

For the domain:

• The expression \(t-1\) must be non-negative; thus, \(t \geq 1\).
• Considering the interval \(0 \leq t \leq 5\), this is reduced to \(1 \leq t \leq 5\).
• Within this range, \(t\) can only take values from 1 to 5, inclusive.

For the range:

• The minimum value happens at \(t=1\), with \(f(t) = \sqrt{1-1} = 0\).
• The maximum value occurs at \(t=5\), with \(f(t) = \sqrt{5-1} = 2\).

Therefore, the function's range, based on the domain \([1,5]\), is \([0,2]\). This provides a complete view of the function's output based on allowed inputs within the defined conditions.

Exploring Square Roots

Square roots are mathematical operations that find a value that, when multiplied by itself, yields the original number. They are fascinating because they bring non-linear behavior to functions, particularly those involving expressions under the square root.

For example, in \(f(t) = \sqrt{t-1}\):

• The operation inside the square root, \(t-1\), determines when the function is "active" or defined.
• When \(t < 1\), \(t-1\) becomes negative, and the square root is not defined for real numbers.

Square roots emphasize restricting domain choices to non-negative values of the expression under the root. This ensures the whole function remains real-valued. As square roots essentially reverse squaring, they can compress and expand the output values in unique ways, which is why determining both domain and range for functions like this is critical. This means understanding square roots helps visualize how adjustments in \(t\) affect \(f(t)\) outcomes directly.

Mastering these concepts can deepen insights into the subject of functions and their applications.