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Question 67976: Could someone please please help me with this:
f(x)= (x²-1)(x-1).
1.Show why f(x)is greater or equal to 0 in each of the following cases:
a)x>1
b)0
c)-1
2.Deduce that x^3+1>x^2+x for all x>-1.
Thank you very for your great help!!!!!!!
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! Could someone please please help me with this:
f(x)= (x²-1)(x-1).
First completely factor the expression for f(x)
f(x) = (x-1)(x+1)(x-1)
f(x) = (x-1)²(x+1)
1.Show why f(x)is greater or equal to 0 in each of the following cases:
a) x > 1
1. x > 1 Given
2. x+1 > 2 Added 1 to both sides of 1
3. (x-1)²(x+1) > 2(x-1)² Multiplied both sides of 2 by
nonnegative quantity (x-1)², which is
nonnegative because the square of any
quantity is nonnegative.
4. f(x) > 2(x-1)² Substituted f(x) for (x-1)²(x+1) in 3.
5. f(x) > 0 Because 4 tells us f(x) is greater than or
equal to twice the square of a quantity, which
is nonnegative.
----------------------------------------------------------
b) 0 < x < 1
1. 0 < x < 1 Given
2. 0+1 < x+1 < 1+1 Added 1 to all 3 sides of 1
3. 1 < x+1 < 2 Simplified 2
4. (x-1)²·1 < (x-1)²(x+1) < (x-1)²·2 Multiplied all 3 sides of 3
by non-negative quantity (x-1)²,
which is known to be non-negative
because any quantity squared is
nonnegative.
5. (x-1)² < (x-1)²(x-1) < 2(x+1)² Simplified 4
6. (x-1)²(x+1) > (x-1)² Writing 5 equivalently. (Don't need
right side of 5.)
7. f(x) > (x-1)² Substituting f(x) for (x-1)²(x+1)
8. f(x) > 0 Because 7 tells us f(x) is greater
than a square quantity, and a
squared quantity is greater than or
equal to 0.
--------------------------------------------
c) -1 < x < 0
1. -1 < x < 0 Given
2. -1+1 < x+1 < 0+1 Added 1 to all 3 sides of 1
3. 0 < x+1 < 1 Simplified 2
4. (x-1)²·0 > (x-1)²(x+1) > (x-1)²·1 Multiplied all 3 sides of 3
by non-negative quantity (x-1)²,
which is known to be non-negative
because any quantity squared is
nonnegative.
5. 0 > (x-1)²(x+1) > (x-1)² Simplified 4
6. (x-1)²(x+1) > 0 Writing 5 equivalently. (Don't need
right side of 5.)
7. f(x) > 0 Substituting f(x) for (x-1)²(x+1)
------
2.Deduce that
x³+1 > x²+x for all x > -1
This isn't true because when x = 1, the inequality
becomes 2 > 2 which is false.
You must have meant
x³+1 > x²+x for all x > -1
Factor both sides of the inequality, to see what
we need to prove. We need to prove this:
(x+1)(x²-x+1) > x(x+1)
1. x > -1 Given
2. x+1 > 0 Added 1 to both sides of 1
Now we need to work backwards and look at what we need.
We need to show that x²-x+1 > x so we can multiply both
sides of 2 by x+1 which we know is positive from 2. But
that inequality is equivalent to x²-2x+1 > 0 by adding
-x to both sides. Then that is equivalent to (x-1)² > 0.
But that is true because any squared quantity is
nonnegative. So now we reverse our thinking here:
3. (x-1)² > 0 The square of any quantity is
nonnegative.
4. x²-2x+1 > 0 Squared x-1 in 3.
5. x²-x+1 > x Added x to both sides.
6. (x+1)(x²-x+1) > x(x+1) Multiplied both sides of 5 by positive
quantity (x+1), which we know is
positive since 2 tells us it is greater
than 0.
That's what we had to prove.
Edwin
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