## Calculus: Early Transcendentals 8th Edition

$f(1) = -0.84$ $f(2) = 1.1$
$sinx = x^{2} - x$ Equal this fuction to $f(x)$. $0= x^{2} - x-sinx$ Now equal the value of $x$ to the values given in the interval: $(1,2)$. $f(1) = -sin(1) + 1^{2} - 1 = -sin(1) = -0.84$ $f(2) = -sin(2) +2^{2} - 2 = sin(2) -2 = 1.1$ The intermediate value theorem takes all the values between $f(1)$ and $f(2)$. So in this case $f(x)$ will take all the values between $-0.84$ and $1.1$.