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Question 64228: Help please!
1.) The solution set of the inequality x+5/x-2<=6 in interval notation
2.) If x+243, then x^-1/3 is ?
3.) The inequality -4x+7<=2x+4 is equivalent to ?
4.) What is the largest zero of the function f(x)=x^3+5x^2+5x-2
Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website! Help please!
1.) The solution set of the inequality x+5/x-2<=6 in interval notation
x + 5
------- £ 6
x - 2
Get 0 on the right by subtracting 6 from both sides:
x + 5
------- - 6 £ 0
x - 2
Write 6 over 1
x + 5 6
------- - --- £ 0
x - 2 1
LCD = x - 2. The first fraction already has that
LCD. The make the second fraction have it too, we
multiply top and bottom by it:
x + 5 6(x - 2)
------- - ---------- £ 0
x - 2 1(x - 2)
x + 5 6x - 12
------- - ---------- £ 0
x - 2 x - 2
Combine the two fractions:
x + 5 - (6x - 12)
------------------- £ 0
x - 2
x + 5 - 6x + 12
------------------ £ 0
x - 2
-5x + 17
---------- £ 0
x - 2
The zero of the numerator is 17/5 or 3.4
The zero of the denominator is 2
So the critical values are 2 and 3.4
We draw a number line and mark those two
values:
-----------o----------o-------------
2 3.4
Select a test value less than 2, say 0,
Substitute it into
-5x + 17
---------- £ 0
x - 2
-5(0) + 17
------------ £ 0
(0) - 2
-17/2 £ 0
This is true, so we shade that part of
the number line
< ===========o----------o-------------
2 3.4
Now choose a test value between 2 and 3.4,
say 3
Substitute it into
-5x + 17
---------- £ 0
x - 2
-5(3) + 17
------------ £ 0
(3) - 2
2 £ 0
This is false, so we do not shade that part of
the number line.
< ===========o----------o-------------
2 3.4
Now choose a test value greater than 3.4,
say 4
Substitute it into
-5x + 17
---------- £ 0
x - 2
-5(4) + 17
------------ £ 0
(4) - 2
-3/2 £ 0
This is true, so we shade that part of
the number line
< ===========o----------o============ >
2 3.4
Now we have to test the critical points themselves.
x cannot be 2 because that causes the denominator
to be 0, which is undefined. However x can be 3.4
because that makes only the numerator 0 and since
the inequality is " £ " and not " < ", we can
include 0. So we darken the circle at 3.4.
< ===========o----------·============ >
2 3.4
So the interval notation for the solution is
(-¥, 2) È [3.4, ¥)
2.) If x=243, then x-1/3 is ?
Notice that 243 = 3·3·3·3·3 = 35
1
x-1/3 = 243-1/3 = (35)-1/3 = ----------- =
(35)1/3
1 1 1 1 1
-------- = -------- = --------- = --------- = -------- =
35/3 31+2/3 31·32/3 3·32/3 3(³Ö3²)
1 ³Ö3 ³Ö3 ³Ö3 ³Ö3
--------·----- = --------- = ----- = ----
3(³Ö3²) ³Ö3 3(³Ö3³) 3·3 9
3.) The inequality -4x+7<=2x+4 is equivalent to ?
Solve it like an equation until the last step:
-4x + 7 < 2x + 4
- 7 - 7
-----------------------
-4x < 2x - 3
-2x -2x
-----------------------
-6x < -3
But when you get to the last step, where you divide
by the coefficient of x, if you divide by a POSITIVE
number you KEEP the inequality sign as is, but if
you divide by a NEGATIVE number you must REVERSE the
inequality sign. Here we must divide by -6, which is
NEGATIVE, so we must change the < to >.
-6x/(-6) > -3/(-6
x > 1/2
4.) What is the largest zero of the function f(x)=x^3+5x^2+5x-2
The only way to find it is to approximate it by using a graphing
calculator. It is approximately 0.3027756377
Edwin
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