#### Answer

The graph of the function has at least two x-intercepts in the interval $(1,2)$

#### Work Step by Step

$y = sin~x^3$
We can see that this function is continuous in the interval $(1,2)$.
Let $x = 1.1$
$y = sin~(1.1)^3 = 0.97$
Let $x = 1.5$
$y = sin~(1.5)^3 = -0.23$
Let $x = 1.9$
$y = sin~(1.9)^3 = 0.54$
By the Intermediate Value Theorem, there is a number $b$ in $(1.1, 1.5)$ such that $sin~b^3 = 0$, and there is a number $c$ in $(1.5,1.9)$ such that $sin~c^3 = 0$
The graph of the function has at least two x-intercepts in the interval $(1,2)$