How do solve #x^2<=8x# and write the answer as a inequality and interval notation?

1 Answer
Dec 16, 2016

Answer:

The answer is #x in [0 ,8 ] # or #0<=x<=8#

Explanation:

Let's rewrite the equation as

#x^2-8x<=0#, #=>#, #x(x-8)<=0#

Let #f(x)=x(x-8)#

Let's do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##0##color(white)(aaaa)##8##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-8##color(white)(aaaaa)##-##color(white)(aaa)##-##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaa)##-##color(white)(aaa)##+#

Therefore,

#f(x)<=0# when #x in [0 ,8 ] #, or #0<=x<=8#