Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

In order to be continuous at \(2\) we require

to hold. We use the three part definition listed previously to check this.

1. First, \(f(2)=c\text{,}\) and \(c\) is some real number. Thus, \(f(2)\) is defined.

2. Now, we must evaluate the limit. Rather than computing both one-sided limits, we just compute the limit directly. For \(x\) close to \(2\) (but not equal to \(2\)) we can replace \(f(x)\) with \(\frac{x^2-x-2}{x-2}\) to get:

   \begin{equation*} \lim_{x\to 2}f(x)=\lim_{x\to 2}\frac{x^2-x-2}{x-2}=\lim_{x\to 2}\frac{(x-2)(x+1)}{x-2}=\lim_{x\to 2}(x+1)=3\text{.} \end{equation*}

   Therefore the limit exists and equals \(3\text{.}\)

3. Finally, for \(f\) to be continuous at \(2\text{,}\) we need that the numbers in the first two items to be equal. Therefore, we require \(c=3\text{.}\)

Thus, when \(c=3\text{,}\) \(f(x)\) is continuous at \(2\text{,}\) for any other value of \(c\text{,}\) \(f(x)\) is discontinuous at \(2\text{.}\)

We will use a continuity argument to justify that direct substitution can be applied. By the list above, \(\sqrt{x}\text{,}\) \(\sin x\text{,}\) \(1\text{,}\) \(x\) and \(\cos x\) are all continuous functions at \(\pi\text{.}\) Then \(\sqrt x+\sin x\) and \(1+x+\cos x\) are both continuous at \(\pi\text{.}\) Finally,

is a continuous function at \(\pi\) since \(1+\pi+\cos\pi\neq 0\text{.}\) Hence, we can directly substitute to get the limit:

1. The functions that make up the composition \(h(x)=f(g(x))\) are \(g(x)=x^2\) and \(f(x)=\cos(x)\text{.}\) The function \(g\) is continuous on \(\mathbb{R}\) since it is a polynomial, and \(f\) is also continuous everywhere. Therefore, \(h(x)=(f\circ g)(x)\) is continuous on \(\mathbb{R}\) by Theorem 3.63.

2. We know from Example 3.61 that \(f(x)=\ln x\) is continuous and \(g(x)=1+\sin x\) are continuous. Thus by Theorem 3.63, \(H(x)=f(g(x))\) is continuous wherever it is defined. Now \(\ln(1+\sin x)\) is defined when \(1+\sin x>0\text{.}\) Recall that \(-1\leq \sin x\leq 1\text{,}\) so \(1+\sin x>0\) except when \(\sin x=-1\text{,}\) which happens when \(x=\pm 3\pi/2,\pm 7\pi/2,\ldots\text{.}\) Therefore, \(H\) has discontinuities when \(x=3\pi n/2\text{,}\) \(n=1,2,3,\ldots\) and is continuous on the intervals between these values.

1. The first question is easy to answer since for any exponential function we know that \(a^x>0\text{,}\) and we also know that whenever you square a number you get a nonnegative answer: \(x^2\geq 0\text{.}\) Hence, \(e^x+x^2>0\text{,}\) and thus, is never equal to zero. Therefore, the first equation has no solution.

2. For the second question, it is difficult to see if \(e^x+x=0\) has a solution. If we tried to solve for \(x\text{,}\) we would run into problems. Let's make a table of values to see what kind of values we get (recall that \(e\approx 2.7183\)):

   \begin{equation*} \begin{array}{l|cccc} x \amp -2 \amp -1 \amp 0 \amp 1 \\ \hline e^x+x \amp e^{-2}-2 \approx -1.9 \amp e^{-1}-1 \approx 1 \amp e^0+0 = 1 \amp e + 1 \approx 3.7 \end{array} \end{equation*}

   Sketching this gives:

   
   Let \(f(x)=e^x+x\text{.}\) Notice that if we choose \(a=-1\) and \(b=0\) then we have \(f(a)\lt 0\) and \(f(b)>0\text{.}\) A point where the function \(f(x)\) crosses the \(x\)-axis gives a solution to \(e^x+x=0\text{.}\) Since \(f(x)=e^x+x\) is continuous (both \(e^x\) and \(x\) are continuous), then the function must cross the \(x\)-axis somewhere between \(-1\) and \(0\text{:}\)
   
   Therefore, our equation has a solution. Note that by looking at smaller and smaller intervals \((a,b)\) with \(f(a)\lt 0\) and \(f(b)>0\text{,}\) we can get a better and better approximation for a solution to \(e^x+x=0\text{.}\) For example, taking the interval \((-0.4,-0.6)\) gives \(f(-0.4)\lt 0\) and \(f(-0.6)>0\text{,}\) thus, there is a solution to \(f(x)=0\) between \(-0.4\) and \(-0.6\text{.}\) It turns out that the solution to \(e^x+x=0\) is \(x\approx -0.56714\text{.}\)

Let \(f(x)=\sqrt[3]{x}+x-1\text{,}\) \(N=0\text{,}\) \(a=0\text{,}\) and \(b=8\text{.}\) Since \(\sqrt[3]{x}\text{,}\) \(x\) and \(-1\) are continuous on \(\mathbb{R}\text{,}\) and the sum of continuous functions is again continuous, we have that \(f(x)\) is continuous on \(\mathbb{R}\text{,}\) thus in particular, \(f(x)\) is continuous on \([0,8]\text{.}\) We have \(f(a)=f(0)=\sqrt[3]{0}+0-1=-1\) and \(f(b)=f(8)=\sqrt[3]{8}+8-1=9\text{.}\) Thus \(N=0\) lies between \(f(a)=-1\) and \(f(b)=9\text{,}\) so the conditions of the Intermediate Value Theorem are satisfied. So, there exists a number \(c\) in \((0,8)\) such that \(f(c)=0\text{.}\) This means that \(c\) satisfies \(\sqrt[3]{c}+c-1=0\text{,}\) in otherwords, is a solution for the equation given.

Alternatively we can let \(f(x)=\sqrt[3]{x}+x\text{,}\) \(N=1\text{,}\) \(a=0\) and \(b=8\text{.}\) Then as before \(f(x)\) is the sum of two continuous functions, so is also continuous everywhere, in particular, continuous on the interval \([0,8]\text{.}\) We have \(f(a)=f(0)=\sqrt[3]{0}+0=0\) and \(f(b)=f(8)=\sqrt[3]{8}+8=10\text{.}\) Thus \(N=1\) lies between \(f(a)=0\) and \(f(b)=10\text{,}\) so the conditions of the Intermediate Value Theorem are satisfied. So, there exists a number \(c\) in \((0,8)\) such that \(f(c)=1\text{.}\) This means that \(c\) satisfies \(\sqrt[3]{c}+c=1\text{,}\) in otherwords, is a solution for the equation given.

By Theorem 3.9, \(f\) is continuous. Since \(f(0)=-2\) and \(f(1)=3\text{,}\) and \(0\) is between \(-2\) and \(3\text{,}\) there is a \(c\in(0,1)\) such that \(f(c)=0\text{.}\)

If we compute \(f(0.1)\text{,}\) \(f(0.2)\text{,}\) and so on, we find that \(f(0.6)\lt 0\) and \(f(0.7)>0\text{,}\) so by the Intermediate Value Theorem, \(f\) has a root between \(0.6\) and \(0.7\text{.}\) Repeating the process with \(f(0.61)\text{,}\) \(f(0.62)\text{,}\) and so on, we find that \(f(0.61)\lt 0\) and \(f(0.62)>0\text{,}\) so \(f\) has a root between \(0.61\) and \(0.62\text{,}\) and the root is \(0.6\) rounded to one decimal place.