Problem 38 - 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 \(37-42\). $$ f(t)=1 / \sqrt{t-1} $$

Step-by-Step Solution

Answer

The domain is \(1

Step 1: Determine the Domain of the Function

The function given is \( f(t) = \frac{1}{\sqrt{t-1}} \). For \(f(t)\) to be defined, the expression under the square root, \(t-1\), must be greater than zero. Therefore, we set the inequality \(t-1>0\). Solving this gives \(t>1\). The domain is all values of \(t\) greater than 1. Since we are given the interval \(t=0\) to \(t=5\), the domain in this range is \(1

Step 2: Determine the Range in 37-42

Next, we need to find the range of \(f(t)\) when its values are close to falling within the interval 37-42. The function \(f(t) = \frac{1}{\sqrt{t-1}} \) decreases as \(t\) increases. Calculate \(f(t)\) for the boundary values of \(t\), namely at points close to \(t=1\) and \(t=5\), and analyze range from there. As \(t \to 1^+\), \(f(t)\to +\infty\), and at \(t=5\), \(f(5)=\frac{1}{\sqrt{4}}=\frac{1}{2}\). Thus, \(f(t)\) takes values from \((+\infty,\frac{1}{2}]\). Since 37-42 is not included in this range, \(f(t)\) never falls into 37-42 for \(1

Key Concepts

Inequalities

Inequalities are essential in determining the domain of a function, especially when dealing with square roots or divisions by an expression. In a function like \(f(t) = \frac{1}{\sqrt{t-1}}\), we need to ensure that the expression inside the square root, \(t-1\), is greater than zero because the square root of a negative number is not defined in the real numbers. To find when \(\sqrt{t-1}\) is valid, you solve the inequality \(t-1 > 0\), leading to \(t > 1\). This calculation tells us where the function is defined. Inequalities help us set boundaries to ensure we only use valid inputs that keep our function real and defined.

Square Root Function

The square root function is a type of radical function, represented generally as \(\sqrt{x}\). These functions are defined only when the expression inside the square root is non-negative, since taking a square root of a negative number is undefined in the real number system. In the function \(f(t) = \frac{1}{\sqrt{t-1}}\), the valid values of \(t\) are those that make the expression under the square root positive. Because we only want positive values under the square root, we set \(t-1 > 0\), leading us to the domain of \(t > 1\). This ensures when \(f(t)\) is calculated, it results in a real, defined value.

Interval Notation

Interval notation is a concise way of describing a set of numbers between two endpoints. It helps in specifying ranges clearly and efficiently. For the function \(f(t)\), the domain based on \(t > 1\) and restricted within the interval given as \(t=0\) to \(t=5\), is written in interval notation as \((1, 5]\). The rounded parenthesis \(( \) symbol tells us that \(1\) is not included (open interval), whereas the square bracket \([ \) shows that \(5\) is included (closed interval). Thus, interval notation can clarify which values a function can take, eliminating any ambiguity.

Function Decreasing Behavior

In understanding how a function behaves over a particular interval, it's important to note its increasing or decreasing nature. For \(f(t) = \frac{1}{\sqrt{t-1}}\), the function is decreasing. This means as \(t\) increases, \(f(t)\) decreases. A decreasing function implies that for any two numbers \(a\) and \(b\) where \(a < b\), the value of \(f(a) > f(b)\). In this context, as \(t\) moves from slightly greater than \(1\) towards \(5\), \(f(t)\) starts from very large values (approaching infinity) and decreases, reaching \(\frac{1}{2}\) when \(t=5\). Understanding this helps to determine why the range for \(f(t)\) only reaches down to \(\frac{1}{2}\) for \(1 < t \leq 5\), and why it doesn’t go lower into the interval \(37-42\).

Previous

Problem 38

Next

Problem 38

Other exercises in this chapter

Problem 37

Which points between \(t=0\) and \(t=5\) can be in the domain of \(f(t)\) ? With this domain find the range in \(37-42\). $$ f(t)=\sqrt{t-1} $$

View solution

Problem 38

Choose the number \(k\) so that \(x=\cos k t, y=\sin k t\) completes a rotation at \(t=1 .\) Find the speed and upward velocity.

View solution

Problem 38

Draw rough graphs or computer graphs of \(t \sin t\) and \(\sin 4 t \sin t\) from 0 to \(2 \pi\).

View solution

Problem 39

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)

View solution

Key Concepts

Inequalities

Inequalities are essential in determining the domain of a function, especially when dealing with square roots or divisions by an expression. In a function like \(f(t) = \frac{1}{\sqrt{t-1}}\), we need to ensure that the expression inside the square root, \(t-1\), is greater than zero because the square root of a negative number is not defined in the real numbers. To find when \(\sqrt{t-1}\) is valid, you solve the inequality \(t-1 > 0\), leading to \(t > 1\). This calculation tells us where the function is defined. Inequalities help us set boundaries to ensure we only use valid inputs that keep our function real and defined.

Square Root Function

The square root function is a type of radical function, represented generally as \(\sqrt{x}\). These functions are defined only when the expression inside the square root is non-negative, since taking a square root of a negative number is undefined in the real number system. In the function \(f(t) = \frac{1}{\sqrt{t-1}}\), the valid values of \(t\) are those that make the expression under the square root positive. Because we only want positive values under the square root, we set \(t-1 > 0\), leading us to the domain of \(t > 1\). This ensures when \(f(t)\) is calculated, it results in a real, defined value.

Interval Notation

Interval notation is a concise way of describing a set of numbers between two endpoints. It helps in specifying ranges clearly and efficiently. For the function \(f(t)\), the domain based on \(t > 1\) and restricted within the interval given as \(t=0\) to \(t=5\), is written in interval notation as \((1, 5]\). The rounded parenthesis \(( \) symbol tells us that \(1\) is not included (open interval), whereas the square bracket \([ \) shows that \(5\) is included (closed interval). Thus, interval notation can clarify which values a function can take, eliminating any ambiguity.

Function Decreasing Behavior

In understanding how a function behaves over a particular interval, it's important to note its increasing or decreasing nature. For \(f(t) = \frac{1}{\sqrt{t-1}}\), the function is decreasing. This means as \(t\) increases, \(f(t)\) decreases. A decreasing function implies that for any two numbers \(a\) and \(b\) where \(a < b\), the value of \(f(a) > f(b)\). In this context, as \(t\) moves from slightly greater than \(1\) towards \(5\), \(f(t)\) starts from very large values (approaching infinity) and decreases, reaching \(\frac{1}{2}\) when \(t=5\). Understanding this helps to determine why the range for \(f(t)\) only reaches down to \(\frac{1}{2}\) for \(1 < t \leq 5\), and why it doesn’t go lower into the interval \(37-42\).