Question 62424: If f(x)=x(x-1)(x-4)^2, use interval notation to give all values of x where f(x)>0.
Found 2 solutions by funmath, Edwin McCravy:
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! If f(x)=x(x-1)(x-4)^2, use interval notation to give all values of x where f(x)>0.
:
Find the places where the f(x)=0, some books call them critical numbers. This will tell you what the possible intervals are.
x=0 and x-1=0 and x-4=0
x=0 and x=1 and x=4
:
The intervals to test are: (-infinity,0), (0,1), (1,4) and (4,infinity)
:
Pick a test number from each inveral, if it's >0 then the interval is part of the soltuion, if not, it's not.
:
For (-infinity,0) test x=-1.
-1(-1-1)(-1-4)^2>0 ?
-1(-2)(-5)^2>0 ?
-1(-2)(25)>0 ? An even number of negatives is +.
50>0 This is true so (-infinity,0) is part of the solution.
:
For (0,1) test x=1/2
1/2(1/2-1)(1/2-4)^2>0 ?
1/2(-1/2)(-7/2)^2>0 ?
1/2(-1/2)(49/4)>0 We don't have to solve this to see that it's negative because an odd number of negatives is -.
(0,1) is rejected.
:
For (1,4) test x=3
3(3-1)(3-4)^2>0
3(2)(-1)^2>0
3(2)(1)>0 No positive means this is positive, therefore (1,4) is part of the solution.
:
For (4,infinity) test x=5
5(5-1)(5-4)^2>0
5(4)(1)^2>0
5(4)(1)>0 No positive means this is positive, therefore (4,infinity) is part of the solution set.
Therefore the solution set is: (-infinity,0)U(1,4)U(4,infinity)
Happy Calculating!!!
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website! If f(x)=x(x-1)(x-4)^2, use interval
notation to give all values of x where
f(x)>0.
Find the critical values by setting all
the factors of numerators and
denominators = 0
x = 0, x-1 = 0, x-4 = 0
gives critical values 0, 1 and 4
Mark these on a number line
覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧
-2 -1 0 1 2 3 4 5 6
Choose any value left of 0, say -1 and find
f(-1)
f(x) = x(x-1)(x-4)ｲ
f(-1) = -1(-1-1)(-1-4)ｲ
f(-1) = -1(-2)(-5)ｲ
f(-1) = +50
We don't care about the value of +50,
but only about the fact that it's positive,
so we place positive signs above the number
line left of 0:
+ + + + +
覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧
-2 -1 0 1 2 3 4 5 6
Next we choose any value of x between 0 and 1,
say .5 and find
f(.5)
f(x) = x(x-1)(x-4)ｲ
f(.5) = .5(.5-1)(.5-4)ｲ
f(-1) = .5(-.5)(-3.5)ｲ
f(-1) = -3.0625
We don't care about the value of -3.0625,
but only about the fact that it's negative,
so we place negative signs above the number
line between 0 and 1:
+ + + + + - -
覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧
-2 -1 0 1 2 3 4 5 6
Next we choose any value of x between 1 and 4,
say 2, and find
f(2)
f(x) = x(x-1)(x-4)ｲ
f(2) = 2(2-1)(2-4)ｲ
f(2) = 2(1)(-2)ｲ
f(2) = +8
We don't care about the value of +8,
but only about the fact that it's positive,
so we place positive signs above the number
line between 1 and 4:
+ + + + + - - + + + + + +
覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧
-2 -1 0 1 2 3 4 5 6
Next we choose any value of x to the right of 4,
say 5, and find
f(5)
f(x) = x(x-1)(x-4)ｲ
f(5) = 5(5-1)(5-4)ｲ
f(2) = 5(4)(1)ｲ
f(2) = +20
We don't care about the value of +20,
but only about the fact that it's positive,
so we place positive signs above the number
line between 1 and 4:
+ + + + + - - + + + + + + + + + +
覧覧覧覧熔覧熔覧覧覧覧覧熔覧覧覧覧
-2 -1 0 1 2 3 4 5 6
Now we notice that the original problem was to
find all values of x for which f(x) > 0, which
means that f(x) is positive, so we are to write
the interval notation for the part of the number
line which has positive signs above it, namely
this part:
+ + + + + - - + + + + + + + + + +
<=========o覧熔===========o========>
-2 -1 0 1 2 3 4 5 6
The solution will not include the critical points
since the inequality is > 0 and not >0.
Answer in interval notation:
(-oo, 0) U (1,4) U (4,oo)
where oo is the infinity symbol.
Edwin
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