#### Answer

1) Prove that the function $f(x)=x^5-x^3+3x-5$ is continuous on its domain.
2) Calculate $f(1)$ and $f(2)$
3) Show that $0\in(f(1),f(2))$. From that, according to the Intermediate Value Theorem, we can show that there must be a number $a$ in the interval $(1,2)$ so that $f(a)=0$. That means there is a root of the equation in interval $(1,2)$.

#### Work Step by Step

$$x^5-x^3+3x-5=0$$ interval $(1,2)$
*STRATEGY:
1) Prove that the function $f(x)=x^5-x^3+3x-5$ is continuous on its domain.
2) Calculate $f(1)$ and $f(2)$
3) Show that $0\in(f(1),f(2))$. From that, according to the Intermediate Value Theorem, we can show that there must be a number $a$ in the interval $(1,2)$ so that $f(a)=0$. That means there is a root of the equation in interval $(1,2)$.
1) Since $f(x)$ is a polynomial function, it is continuous at any number in its domain, which is $R$.
Therefore, the interval $(1,2)$ lies in the domain of $f(x)$.
2) Calculate:
$f(1)=1^5-1^3+3\times1-5=1-1+3-5=-2$
$f(2)=2^5-2^3+3\times2-5=25$
3) Since $f(1)\lt0$ and $f(2)\gt0$, $0$ lies in the interval $(f(1), f(2))$.
According to the Intermediate Value Theorem, since $f(x)$ is continuous on $(1,2)$, there must a number $a$ in $(1,2)$ that $f(a)=0$.
That means there must be a root of the equation in the interval $(1,2)$.