함수의 극한과 연속 전체 문제

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 $c=1$.