Which shows all the critical points for the inequality x^2/9x/x-5<0

First, you need to have an inequality and you have not writen an inequality, so I am going to do an example with an actual inequality bases of some expressions similar to those of the question:

If the ineqquality is (x^2 - 9x ) / (x - 5) > 0

The critical points are where x^ - 9x = 0 and where x - 5 = 0

1) x^2 - 9x = 0 => x (x - 9) = 0 => x = 0 and x = 9

2) x - 5 = 0 => x = 5

So, for this case the critical points would be 0, 5 and 9.

Hope I helped! :)

MrRoyal MrRoyal · Answer 2 · 2021-11-04T13:50:28+01:00

The critical point is a single value at which the function is not differentiable

The critical points are 0, 5 and 9

The inequality is given as:

[tex]\mathbf{\frac{x^2 - 9x}{x -5} < 0}[/tex]

Factorize the numerator

[tex]\mathbf{\frac{x(x - 9)}{x -5} < 0}[/tex]

Equate the numerator and the denominator to 0

[tex]\mathbf{x(x - 9) = 0}[/tex] and [tex]\mathbf{x -5=0}[/tex]

Split

[tex]\mathbf{x = 0}[/tex] , [tex]\mathbf{x - 9= 0}[/tex] and [tex]\mathbf{x - 5= 0}[/tex]

Solve for x

[tex]\mathbf{x = 0}[/tex] , [tex]\mathbf{x = 9}[/tex] and [tex]\mathbf{x = 5}[/tex]

Hence, the critical points are 0, 5 and 9