Question 135437
*[Tex \LARGE x^2-8x+12\le0]  Start with the given inequality



*[Tex \LARGE (x-2)(x-6)\le0]  Factor the left side



*[Tex \LARGE (x-2)(x-6)=0]  Set the left side equal to zero



*[Tex \LARGE  x=2  \textrm{ or }   x=6  ] Solve for x in each case



So this means that the critical values are



*[Tex \LARGE  x=2  \textrm{ or }   x=6  ]



Now plot the critical values on a number line


{{{number_line( 600, -10, 10, 2,6 )}}}




So let's evaluate a point that is to the left of *[Tex \LARGE  x=2 ] (which is the left most endpoint). Let's evaluate the value {{{x=1}}}

  

*[Tex \LARGE (x-2)(x-6)\le0]  Start with the given inequality


*[Tex \LARGE (1-2)(1-6)\le0]  Plug in {{{x=1}}}


*[Tex \LARGE 5\le 0]  Evaluate and simplify



 Since *[Tex \LARGE 5\le 0] is false, this means that the interval does not work. So that means that this interval is not in the solution set and can be ignored.

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Now let's test a point that is in between the critical values *[Tex \LARGE  x=2 ] and *[Tex \LARGE  x=6 ]



*[Tex \LARGE (x-2)(x-6)\le0]  Start with the given inequality


*[Tex \LARGE (4-2)(4-6)\le0]  Plug in {{{x=4}}}


*[Tex \LARGE -4\le 0]  Evaluate and simplify


So the graph now looks like 



{{{drawing(500,80,-4, 16,-10, 10,
number_line( 500, -4, 16),


line(2,0,6,0),
line(2,0.3,6,0.3),
line(2,0.15,6,0.15),
line(2,-0.15,6,-0.15),
line(2,-0.3,6,-0.3),

circle(6,0,0.2),
circle(6,0,0.15),
circle(6,0,0.1),
circle(6,0,0.2-0.02),

circle(2,0,0.2),
circle(2,0,0.15),
circle(2,0,0.1),
circle(2,0,0.2-0.02)
)}}}



Since *[Tex \LARGE -4\le 0] is true, this means that one part of the solution in interval notation is <font size="8">[</font>*[Tex \LARGE 2,6]<font size="8">]</font>





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So let's evaluate a point that is to the right of *[Tex \LARGE  x=6 ] (which is the right most endpoint). Let's evaluate the value {{{x=7}}}

  

*[Tex \LARGE (x-2)(x-6)\le0]  Start with the given inequality


*[Tex \LARGE (7-2)(7-6)\le0]  Plug in {{{x=7}}}


*[Tex \LARGE 5\le 0]  Evaluate and simplify



 Since *[Tex \LARGE 5\le 0] is false, this means that the interval does not work. So that means that this interval is not in the solution set and can be ignored.

-----------------------------------------------



Answer:


So the solution in interval notation is 

<font size="8">[</font>*[Tex \LARGE 2,6]<font size="8">]</font>