How do you find g(1/3) given $g (x) = x^{2} - x$ $g(x)=x^2-x$ ?

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Answer 1 · 2016-12-19T08:46:23

$- \frac{2}{9}$ $-2/9$

$g (x) = x^{2} - x$ $g(x)=x^2-x$

$g (\frac{1}{3})$ $g(1/3)" "$ means we replace all the $x s$ $""xs" "$ in the equation with $\frac{1}{3}$ $" "1/3""$

$g (\frac{1}{3}) = {(\frac{1}{3})}^{2} - (\frac{1}{3})$ $g(1/3)=(1/3)^2-(1/3)$

$g (x) = \frac{1}{9} - \frac{1}{3} = \frac{1}{9} - \frac{3}{9} = - \frac{2}{9}$ $g(x)=1/9-1/3=1/9-3/9=-2/9$

How do you find g(1/3) given g(x)=x2−xg(x)=x^2-x?