Answer
Use the Intermediate Value Theorem to show that there exists values of $c$ such that $f(c)$ equals the given values.
Work Step by Step
$$f(x)=x^3-8x+10$$
- Domain: $(-\infty,\infty)$
1) Continuity:
On $(-\infty,\infty)$, we notice that $\lim_{x\to c}(x^3-8 x+10)=c^3-8c+10=f(c)$
That means $f$ is continuous on $(-\infty,\infty)$.
2) Show that $c$ exists which $f(c)=\pi$
Here I graph $f(x)=x^3-8 x+10$ for better findings, which I have also enclosed below.
- At $x=0$: $f(0)=0^3-8(0)+10=10$
- At $x=2$: $f(2)=2^3-8(2)+10=8-16+10=2$
Here, since $2\lt\pi\lt10$ and $f$ is continuous on $[0,2]$, according to Intermediate Value Theorem, there exists a value of $x=c\in[0,2]$ such that $f(c)=\pi$.
3) Show that $c$ exists which $f(c)=-\sqrt3$
- At $x=0$: $f(0)=0^3-8(0)+10=10$
- At $x=-4$: $f(-4)=(-4)^3-8(-4)+10=-64+32+10=-22$
Here, since $-22\lt-\sqrt3\lt10$ and $f$ is continuous on $[-4,0]$, according to Intermediate Value Theorem, there exists a value of $x=c\in[-4,0]$ such that $f(c)=-\sqrt3$.
4) Show that $c$ exists which $f(c)=5,000,000$
- At $x=0$: $f(0)=0^3-8(0)+10=10$
- At $x=1000$: $f(1000)=1000^3-8(1000)+10=999,992,010$
Here, since $10\lt5,000,000\lt999,992,010$ and $f$ is continuous on $[0,1000]$, according to Intermediate Value Theorem, there exists a value of $x=c\in[0,1000]$ such that $f(c)=5,000,000$.
