Answer
$f(1)\approx0.84$
$f(2)\approx-1.09$
There is a root of the equation $sin~ x − x^2 + x = 0.$ or
$sin~ x=x^2 − x$. in the interval $(1, 2)$.
Work Step by Step
Given:
$sin~ x=x^2 − x$.
The given equation $sin~ x=x^2 − x$ is equivalent to the equation $sin~ x − x^2 + x = 0.$ We can write the equation in function form as $f(x)=sin~ x − x^2 + x$, which is continous on the
interval $[1, 2]$,
$f(1) =sin~ 1− 1^2 + 1$
$f(1) =sin~ 1− 1 + 1$
$= sin1 − 0 $
$f(1)=sin1$,
$f(1)\approx0.84$
and
$f(2) =sin~2 -2^2 +2$
$ =sin~2 -4+2$
$=sin~2 -2$
$= 0.034-2$
$f(2)\approx -1.09.$
Since $sin~1<0
