How to solve this problem: x^(2) - 6x + 8 ≤ 0?

x^(2) - 6x + 8 ≤ 0
(x - 4)(x - 2) ≤ 0
Solution:
2 ≤ x ≤ 4

x^2 -6x + 8 = (x-2)(x-4)...
f'(x) = 2x - 6
f'(x) = 0 when x = 3
f''(x) = 2, so f(x) is at a minimum when x = 3
So, f(x) </= 0 between x = 2 and x = 4
so
x^2 -6x + 8 </= 0
when 2 </= x </= 4

Draw the graph, and you will find that the function is at or under the x-axis for this range of x

x² - 6x + 8 ≤ 0

Less than or equal to zero, which means you want the left side to resolve to zero or a negative number.

Let's factor the left side:

(x - 4)(x - 2) ≤ 0

So from this, we see that 2 and 4 are roots. Now we just need to figure out what gives us a negative number:

If x > 4, you have positive times positive, which is positive. Not in our solution set.
If 2 < x < 4, you have positive times negative, which is negative. This is in our solution set.
If x < 2, you have negative times negative, which is positive. Not in our solution set.

So including the zeroes since this can equal zero, the answer to this inequality is the range :

2 ≤ x ≤ 4

[ x - 4 ] [ x - 2 ] ≤ 0

____________2_________4_______
x - 4____-ve______-ve___ 0___+ve___
x - 2____-ve__ 0___+ve_______+ve_
product__+ve__0__-ve____0___+ve__

sol set { 2 ≤ x ≤ 4 }

2^2 - 6(2) +8 =0

3^2 - 6(3) +8 = -1

4^2 -6(4) + 8 = 0