Answer
Because f(1)$\lt$0 and f(2)$\gt$0, there exists a value "c" such that f(c) = 0 on the interval.
Work Step by Step
First, you want to set the equation equal to zero.
$x^{2}$-sinx-x=0
From the given interval, you know which points to test, Because the interval is (1,2), you will test the x values of 1 and 2.
f(1) = negative integer
f(2) = positive integer
Therefore, because f(1)$\lt$0 and f(2)$\gt$0, there exists a value "c" such that f(c) = 0 on the interval.
