Solve
(x - 3)(x – 4)/(x – 5)(x – 6)^2 >or= 0
any help?
undefined for x = 5 or x = 6
(x-6)^2 always + where defined
function is zero at x = 3, x = 4
NOW sketch a graph
for x between 3 and 4. the denominator is - and the numerator is -, so >/= 0
between
x = 3 and x = 4
for x less than 3, denominator still - and numerator + so less than 0, not here
for x between 4 and 5, denominator is - and numerator is + so less than 0, not here
for x between 5 and 6, denominator is + and numerator is +, so yes
for x > 6, both numerator and denominator are + so yes