함수의 연속

False. It is continuous over $(-\infty ,0)\cup (0,\infty ).$

For all values of $a,f(a)$ is defined, $\lim_{\theta \to a}f(\theta )$ exists, and $\lim_{\theta \to a}f(\theta )=f(a).$ Therefore, $f(\theta )$ is continuous everywhere.

Nowhere

Since $f$ is a polynomial, it is continuous over all real numbers. $f(0)=1<12345$ and $f(12345)=12345^3+12345^2+12345+1>12345$ (since all terms are positive). So by the IVT, $f(c)=12345$ for some $c$ between $0$ and $12345$.

The function is continuous for $x\ne c$ since each of those two branches are polynomials. So, the only question is whether the function is continuous at $x=c$; for this we need $$\lim_{x\to c^-}f(x)=f(c)=\lim_{x\to c+}f(x).$$ We compute $$\lim_{x\to c^-}f(x)=\lim_{x\to c^-}8-cx = 8-c^2;$$ $$f(c)=8-c\cdot c= 8-c^2\text{ and}$$ $$\lim_{x\to c^+}f(x)=\lim_{x\to c^+}x^2=c^2.$$ So, we need $8-c^2=c^2$, which yields $c^2=4$, i.e. $c=-2$ or $c=2$.

The function is continuous for $x \ne 0$ since $x^2+c$ and $\cos cx$ are continuous everywhere. It remains to check continuity at $x=0$. To do this we must check that the following three are equal.

$$\begin{aligned} \lim_{x \to 0^+} f(x) &= \lim_{x\to 0^+} x^2+c = c\\ f(0) &= c \\ \lim_{x \to 0^-} f(x) &= \lim_{x\to 0^-} \cos cx = \cos 0 = 1 \end{aligned}$$

Hence when $c=1$ we have the limits agree.

The function is continuous for $x\ne c$ since each of those two branches are defined by polynomials. Thus, the only question is whether the function is continuous at $x=c$. Furthermore, $$\lim_{x\to c^-}f(x) = c^2-4$$ and $$\lim_{x\to c^+}f(x) = f(c) = 3c\,.$$ For continunity we need both limits and the value to agree, so $f$ is continuous if and only if $c^2-4 = 3c$, that is if and only if $$c^2-3c-4 = 0\,.$$ Factoring this as $(c-4)(c+1) = 0$ yields $c=-1$ or $c=+4$.

The function is continuous for $x\ne 2c$ since each of those two branches are polynomials. So, the only question is whether the function is continuous at $x=2c$; for this we need $$\lim_{x\to 2c^-}f(x)=f(2c)=\lim_{x\to 2c+}f(x).$$ We compute $$\lim_{x\to 2c^-}f(x)=\lim_{x\to 2c^-}6-cx = 6-2c^2;$$ $$f(2c)=6-c\cdot 2c= 6-2c^2\text{ and}$$ $$\lim_{x\to 2c^+}f(x)=\lim_{x\to 2c^+}x^2=4c^2.$$ So, we need $6-2c^2=4c^2$, which yields $c^2=1$, i.e. $c=-1$ or $c=1$.

This isn't the kind of equality that we can just solve; we'll need a trick, and that trick is the IVT. The general idea is to show that $\sin x$ is somewhere bigger, and somewhere smaller, than $x-1$. However, since the IVT can only show us that a function is equal to a constant, we need to slightly adjust our language. Showing $\sin x = x-1$ is equivalent to showing $\sin x - x + 1 = 0$, so let $f(x)=\sin x - x +1$, and let's show that it has a real root.

First, we need to note that $f(x)$ is continuous (otherwise we can't use the IVT). Now, we need to find a value of $x$ for which it is positive, and for which it's negative. By checking a few values, we find $f(0)$ is positive, and $f(100)$ is negative. So, by the IVT, there exists a value of $x$ (between $0$ and $100$) for which $f(x)=0$. Therefore, there exists a value of $x$ for which $\sin x = x-1$.

We let $f(x)=3^x-x^2$. Then $f(x)$ is a continuous function, since both $3^x$ and $x^2$ are continuous for all real numbers.

We find a value $a$ such that $f(a)>0$. We observe immediately that $a=0$ works since $$f(0)=3^0-0=1>0.$$

We find a value $b$ such that $f(b)<0$. We see that $b=-1$ works since $$f(-1)=\frac{1}{3}-1<0.$$

So, because $f(x)$ is continuous on $(-\infty, \infty)$ and $f(0)>0$ while $f(-1)<0$, then the Intermediate Value Theorem guarantees the existence of a real number $c$ in the interval $(-1,0)$ such that $f(c)=0$.

We let $f(x)=2\tan(x)-x-1$. Then $f(x)$ is a continuous function on the interval $(-\pi/2, \pi/2)$ since $\tan(x)=\sin(x)/\cos(x)$ is continuous on this interval, while $x+1$ is a polynomial and therefore continuous for all real numbers.

We find a value $a\in (-\pi/2,\pi/2)$ such that $f(a)<0$. We observe immediately that $a=0$ works since $$f(0)=2\tan(0)-0-1=0-1=-1<0.$$

We find a value $b\in (-\pi/2, \pi/2)$ such that $f(b)>0$. We see that $b=\pi/4$ works since $$f(\pi/4)=2\tan(\pi/4) -\pi/4 - 1=2-\pi/4 - 1=1-\pi/4=(4-\pi)/4>0,$$ because $3<\pi<4$.

So, because $f(x)$ is continuous on $[0,\pi/4]$ and $f(0)<0$ while $f(\pi/4)>0$, then the Intermediate Value Theorem guarantees the existence of a real number $c\in (0,\pi/4)$ such that $f(c)=0$.

Let $f(x) = \sqrt{\cos(\pi x)} - \sin(2\pi x) -1/2$. This function is continuous provided $\cos(\pi x)\geq 0$. This is true for $0 \leq x \leq \frac{1}{2}$.

Now $f$ takes positive values on $[0,1/2]$:

$$\begin{aligned} f(0) &= \sqrt{\cos(0)} - \sin(0) -1/2 = \sqrt{1} -1/2 = 1/2. \end{aligned}$$

And $f$ takes negative values on $[0,1/2]$:

$$\begin{aligned} f(1/2) &= \sqrt{\cos(\pi/2)}-\sin(\pi)-1/2 = 0-0-1/2 = -1/2 \end{aligned}$$

(Notice that $f(1/3)=(\sqrt{2}-\sqrt{3})/2-1/2$ also works)

So, because $f(x)$ is continuous on $[0,1/2)$ and $f(0)>0$ while $f(1/2)<0$, then the Intermediate Value Theorem guarantees the existence of a real number $c\in (0,1/2)$ such that $f(c)=0$.

We let $f(x)=\dfrac{1}{\cos^2(\pi x)}-x-\dfrac{3}{2}$. Then $f(x)$ is a continuous function on the interval $(-1/2, 1/2)$ since $\cos x$ is continuous everywhere and non-zero on that interval.

The function $f$ takes negative values. For example, when $x=0$:

$$\begin{aligned} f(0) &= \frac{1}{\cos^2(0)} - 0 - \frac{3}{2} = 1-\frac{3}{2} = -\frac{1}{2} <0. \end{aligned}$$

It also takes positive values, for instance when $x=1/4$:

$$\begin{aligned} f(1/4) &= \frac{1}{(\cos \pi/4)^2} - \frac{1}{4} - \frac{3}{2} \\ &= \frac{1}{1/2} - \frac{1+6}{4} \\ &= 2 - 7/4 = 1/4 >0. \end{aligned}$$

By the IVT there is $c$, $0<c<1/4$ such that $f(c)=0$, in which case

$$\begin{aligned} \dfrac{1}{(\cos\pi c)^2} = c+\dfrac{3}{2}. \end{aligned}$$

• If $f(a)=g(a)$, or $f(b)=g(b)$, then we simply take $c=a$ or $c=b$.
• Suppose $f(a) \neq g(a)$ and $f(b) \neq g(b)$. Then $f(a)<g(a)$ and $g(b)<f(b)$, so if we define $h(x)=f(x)-g(x)$, then $h(a)<0$ and $h(b)>0$. Since $h$ is the difference of two functions that are continuous over $[a,b]$, also $h$ is continuous over $[a,b]$. So, by the Intermediate Value Theorem, there exists some $c \in (a,b)$ with $h(c)=0$; that is, $f(c)=g(c)$.