Answer
(a.) Minimum.
(b.) At $x=\frac{1}{2}$, minimum value is $-\frac{3}{2}$.
(c.) Domain $=(-\infty,\infty)$.
Range $=\left[-\frac{3}{2},\infty\right)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=6x^2-6x$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
$a=6,b=-6$ and $c=0$
(a.)
Because $a>0$, the function has a minimum value.
(b.)
$x-$coordinate at which minimum value occurs is
$x=-\frac{b}{2a}$.
Substitute all values.
$x=-\frac{(-6)}{2(6)}$.
Simplify.
$x=\frac{1}{2}$.
Substitute the value of $x$ into the given function.
$f\left(\frac{1}{2}\right))=6\left(\frac{1}{2}\right)^2-6\left(\frac{1}{2}\right)$
Simplify.
$f(\frac{1}{2})=\frac{6}{4}-\frac{6}{2}$
$f\left(\frac{1}{2}\right)=\frac{3}{2}-\frac{6}{2}$
Add numerators because denominators are same.
$f\left(\frac{1}{2}\right)=\frac{3-6}{2}$
Simplify.
$f\left(\frac{1}{2}\right)=-\frac{3}{2}$
(c.)
Domain is all possible input values.
Domain $=(-\infty,\infty)$.
Range is all possible output values.
Minimum value is $-\frac{3}{2}$.
Range $=\left[-\frac{3}{2},\infty\right)$.
