For \(f(x)=\sin(x)\) we recall that the sine function is defined for every possible value of \(x\) but the output is strictly between \(y=-1\) and \(y=1\text{.}\) Therefore, the domain for \(f(x) = \sin(x)\) is \(-\infty \lt x \lt \infty\) and the range is \(-1 \le y \le 1\text{.}\) See the left plot in Figure 5.

For \(g(x) = \sqrt{x}\) we recall that the square root of a negative number results in an imaginary number. In this text we are interested in real-valued output for functions so we must omit all of the negative numbers from the domain and hence \(0 \le x \lt \infty\text{.}\) For the range we recall that the square root of a number will always be a non-negative number. As such, the range is \(0 \le y \lt \infty\text{.}\) See the middle plot in Figure 5.

For \(h(x) = \frac{1}{x}\) we recall that division by zero is mathematically impossible. That is the only troublesome point in the domain so \(-\infty \lt x \lt 0\) or \(0 \lt x \lt \infty\text{.}\) A moment's reflection also reveals that it is impossible to get zero out of the function \(h(x)\) but it is possible to get any other number. Hence \(-\infty \lt y \lt 0\) or \(0 \lt y \lt \infty\text{.}\) See the right plot in Figure 5.

1. The domain of \(f(x)\) is \(0 \le x \le 4\).

2. The approximate range of \(f(x)\) is \(-7 \le y \le 3\). Without the function itself we cannot be sure of the actual heights of the function at the maximum and the minimum.

3. \(f(0) = 0\), \(f(1) = 3\), \(f(3) = -6\), \(f(4) = 0\), and \(f(5)\) does not exist since \(5\) is not in the domain of \(f(x)\).

1. The slope is \(m = \frac{\Delta y}{\Delta x} = \frac{5-(-1)}{(-2)-10} = \frac{6}{-12} = -\frac{1}{2}\). Hence, \begin{equation*} y - 5 = -\frac{1}{2} \left( x-(-2) \right) \end{equation*} which can be rewritten as \begin{equation*} y = -\frac{1}{2} \left( x+2 \right) + 5. \end{equation*}

2. Using the point-slope form of the line we get \begin{equation*} y - 3 = \frac{3}{5} \left( x-2 \right) \end{equation*} which can be rearranged to \begin{equation*} y = \frac{3}{5} \left( x-2 \right) + 3. \end{equation*}

3. Using the slope-intercept form of the line we get \begin{equation*} y = -\frac{2}{3} x - 1. \end{equation*}

First we calculate the slope

Since we have two points and neither is the \(y\) intercept of the linear function we choose to use the point-slope form of the line. Letting \((x_1,y_1) = (5,7)\) we see that

is one form of the linear function. It is often conventient to solve for \(y\) giving us

Notice that we do not necessarily need to simplify all the way to the slope-intercept form of the line.

The common differences can be found for each successive \(y\)-values

The successive differences are clearly the same throughout the data set and the slope for this data set is \(m=5.3\text{.}\) Picking any convenient point, say \((5,12.2)\text{,}\) then allows us to write the equation of the line as

This could be simplified to point-slope form, but there is typically no need for this algebraic simplification.

1. The slope is \begin{equation*} m = \frac{196-203}{700-650} = -\frac{7}{50} \end{equation*} and this slope is consistent no matter which pairs of points we choose.

2. Let \(A\) be the number of occupied apartments and let \(R\) be the rent. The linear function relating the number of occupied apartments and the rent is \begin{equation*} A - 203 = -\frac{7}{50} \left( R - 650 \right). \end{equation*} Solving for \(A\) we get \begin{equation*} A(R) = -\frac{7}{50} \left( R - 650 \right) + 203. \end{equation*} This is a perfectly acceptable algebraic form for the answer, but if you insist on simplifying then \begin{equation*} A(R) = -\frac{7}{50} R + 294. \end{equation*}

3. \(A(1000) = -\frac{7}{50} \left( 1000 - 650 \right) + 203 = 154\) units occupied.

4. The net revenue is the product of the monthly rent and the number of units occupied at that rent. In this case the revenue is \( \$ 1000 \cdot 154 = \$ 154,000\).

1. If \(C = 120\) chirps/minute, substitute this into the function \(T(C)\) to obtain

   \begin{equation*} T(120) = \frac{120}{4} + 40 = 30 + 40 = 70^\circ F. \end{equation*}

2. To find the domain we need to find the appropriate values of \(C\) for the \(T(C)\) function. Solve \(56=C/4+40\) and get \(C = 64\text{.}\) Solve \(136=C/4+40\) and get \(C = 384\text{.}\) So the domain of \(T(C)\) is \(64 \le \text{ chirps/minute } \le 384\) or, in interval notation, \([64, 384]\text{.}\)

1. The slope is \(m = \frac{1}{2}\) and you have a point so use the point-slope form of the line to get \begin{equation*} y - 4 = \frac{1}{2} \left( x-3 \right). \end{equation*} Solving for \(y\) we get \begin{equation*} y = \frac{1}{2} \left( x-3 \right) + 4. \end{equation*}

2. The slope is \(m = -2\) and we have a point so use the point-slope form of the line to get \begin{equation*} y - 4 = -2 (x-3). \end{equation*} Solving for \(y\) we get \begin{equation*} y = -2(x-3) + 4. \end{equation*}

3. The slope is \(m = \frac{1}{3}\) and since we have the \(y\)-intercept we know that the slope-intercept for the line is the proper choice. Hence, \begin{equation*} y = \frac{1}{3} x - 3. \end{equation*}