## University Calculus: Early Transcendentals (3rd Edition)

Use the Intermediate Value Theorem to show that there exists values of $c$ such that $f(c)$ equals the given values.
$$f(x)=x^3-8x+10$$ - 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 $f(c)=5,000,000$.