We proceed by examining the three criteria for continuity.

1. The limits \(\lim\limits_{x\to c} f(x)\) exists for all \(c\) between \(0\) and \(3\text{.}\)

2. \(f(c)\) is defined for all \(c\) between \(0\) and \(3\text{,}\) except for \(c=1\text{.}\) We know immediately that \(f\) cannot be continuous at \(x=1\text{.}\)

3. The limit \(\lim\limits_{x\to c} f(x) = f(c)\) for all \(c\) between \(0\) and \(3\text{,}\) except, of course, for \(c=1\text{.}\)

We conclude that \(f\) is continuous at every point of the interval \((0,3)\) except at \(x=1\text{.}\) Therefore \(f\) is continuous on \((0,1)\cup(1,3)\text{.}\)

We examine the three criteria for continuity.

1. The limits \(\lim\limits_{x\to c} f(x)\) do not exist at the jumps from one “step” to the next, which occur at all integer values of \(c\text{.}\) Therefore the limits exist for all \(c\) except when \(c\) is an integer.

2. The function is defined for all values of \(c\text{.}\)

3. The limit \(\lim\limits_{x\to c} f(x) = f(c)\) for all values of \(c\) where the limit exist, since each step consists of just a line.

We conclude that \(f\) is continuous everywhere except at integer values of \(c\text{.}\) So the intervals on which \(f\) is continuous are \begin{equation*} \ldots, (-2,-1), (-1,0), (0,1), (1,2), \ldots. \end{equation*} We could also say that \(f\) is continuous on all intervals of the form \((n,n+1)\) where \(n\) is an integer.

We examine each in turn.

1. The domain of \(f(x) = 1/x\) is \((-\infty,0) \cup (0,\infty)\text{.}\) As it is a rational function, we apply Theorem 1.3.3 to recognize that \(f\) is continuous on all of its domain.

2. The domain of \(f(x) = \sin(x)\) is all real numbers, or \((-\infty,\infty)\text{.}\) Applying Theorem 1.3.5 shows that \(\sin(x)\) is continuous everywhere.

3. The domain of \(f(x) = \sqrt{x}\) is \([0,\infty)\text{.}\) Applying Theorem 1.3.5 shows that \(f(x) = \sqrt{x}\) is continuous on its domain of \([0,\infty)\text{.}\)

4. The domain of \(f(x) = \sqrt{1-x^2}\) is \([-1,1]\text{.}\) Applying Theorems 1.3.1 and 1.3.5 shows that \(f\) is continuous on all of its domain, \([-1,1]\text{.}\)

5. The domain of \(f(x) = \abs{x}\) is \((-\infty,\infty)\text{.}\) We can define the absolute value function as \(f(x) = \begin{cases}-x \amp x\lt 0 \\ x \amp x\geq 0\end{cases}\text{.}\) Each “piece” of this piecewise defined function is continuous on all of its domain, giving that \(f\) is continuous on \((-\infty,0)\) and \([0,\infty)\text{.}\) We cannot assume this implies that \(f\) is continuous on \((-\infty,\infty)\text{;}\) we need to check that \(\lim\limits_{x\to 0}f(x) = f(0)\text{,}\) as \(x=0\) is the point where \(f\) transitions from one “piece” of its definition to the other. It is easy to verify that this is indeed true, hence we conclude that \(f(x) = \abs{x}\) is continuous everywhere.

We examine each in turn, applying Theorems 1.5.8 and 1.5.9 as appropriate.

1. The square root terms are continuous on the intervals \([1,\infty)\) and \((-\infty,5]\text{,}\) respectively. As \(f\) is continuous only where each term is continuous, \(f\) is continuous on \([1,5]\text{,}\) the intersection of these two intervals. A graph of \(f\) is given in Figure 1.5.11.

2. The functions \(y=x\) and \(y=\sin(x)\) are each continuous everywhere, hence their product is, too.

3. Theorem 1.5.9 states that \(f(x) = \tan(x)\) is continuous on its domain. Its domain includes all real numbers except odd multiples of \(\pi/2\text{.}\) Thus \(f(x) = \tan(x)\) is continuous on \begin{equation*} \ldots,\left(-\frac{3\pi}{2},-\frac{\pi}2\right),\left(-\frac{\pi}2,\frac{\pi}2\right),\left(\frac{\pi}2,\frac{3\pi}2\right),\ldots \end{equation*} or, equivalently, on \(D = \left\{x\in \mathbb{R}\mid x\neq (2n+1)\cdot \frac{\pi}2,\text{ where }n \text{ is an integer } \right\}\text{.}\)

4. The range of \(y=\ln(x)\) is \((-\infty,\infty)\) which includes invalid input for \(\sqrt{\phantom{x}}\text{.}\) We must restrict the domain to \(I=[1,\infty)\) in order to have \(\ln(x)\) produce output in \(J=[0,\infty)\text{,}\) which can then be used as valid input into \(\sqrt{\phantom{x}}\text{.}\) Since \(\ln(\phantom{x})\) is continuous on \(I\) and \(\sqrt{\phantom{x}}\) is continuous on \(J\text{,}\) \(f(x) = \sqrt{\ln(x)}\) is continuos on \(I=[1,\infty)\text{.}\)

Consider the graph of \(f(x) = x-\cos(x)\text{,}\) shown in Figure 1.5.15. It is clear that the graph crosses the \(x\)-axis somewhere near \(x=0.8\text{.}\) To start the Bisection Method, pick an interval that contains \(0.8\text{.}\) We choose \([0.7,0.9]\text{.}\) Note that all we care about are signs of \(f(x)\text{,}\) not their actual value, so this is all we display.

Iteration 1:

\(f(0.7) \lt 0\text{,}\) \(f(0.9) > 0\text{,}\) and \(f(0.8) >0\text{.}\) So replace \(0.9\) with \(0.8\) and repeat.

Iteration 2:

\(f(0.7)\lt 0\text{,}\) \(f(0.8) > 0\text{,}\) and at the midpoint, \(0.75\text{,}\) we have \(f(0.75) >0\text{.}\) So replace \(0.8\) with \(0.75\) and repeat. Note that we don't need to continue to check the endpoints, just the midpoint. Thus we put the rest of the iterations in Table 1.5.16.

Notice that in the 12th iteration we have the endpoints of the interval each starting with \(0.739\text{.}\) Thus we have narrowed the zero down to an accuracy of the first three places after the decimal. Using a computer, we have \begin{equation*} f(0.7390) = -0.00014, f(0.7391) = 0.000024. \end{equation*}

Either endpoint of the interval gives a good approximation of where \(f\) is \(0\text{.}\) The 1.5.13 states that the actual zero is still within this interval. While we do not know its exact value, we know it starts with \(0.739\text{.}\)

This type of exercise is rarely done by hand. Rather, it is simple to program a computer to run such an algorithm and stop when the endpoints differ by a preset small amount. One of the authors did write such a program and found the zero of \(f\) to be \(0.7390851332\) , accurate to \(10\) places after the decimal. While it took a few minutes to write the program, it took less than a thousandth of a second for the program to run the necessary \(35\) iterations. In less than \(8\) hundredths of a second, the zero was calculated to \(100\) decimal places (with less than \(200\) iterations).