### Presentation

```Solving Inequalities
By: Sam Milkey and Noah Bakunowicz
Polynomial Inequalities
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A polynomial inequality takes the form f(x) > 0, f(x) ≥ 0,
f(x) < 0, f(x) ≤ 0, or f(x) ≠ 0.
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To solve f(x) > 0 is to find the values of x that make
f(x) positive.
To solve f(x) < 0 is to find the values of x that make
f(x) negative.
But that’s pretty boring.
Example 1 Finding negative, positive,
zero
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F(x)=(x+2)(x+1)(x-5)
Zeros: -2 (mult of 1), -1(mult of 1), 5 (mult of 1)
Number line:
--+-+++++
- -2
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+
-1
-
5
+
Find when it is Zero, Negative, and Positive
o Zeros: -2, -1, 5
o Negative: (∞, -2) (-1, 5)
o Positive: (-2,-1) (5,∞)
Example 2 Solving Algebraically
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Solve 2x³-7x²-10x+24>0 Analytically
Use the rational zeros theorem to find possible rational
zeros
o ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ±1/2, ±3/2
You can use a graph to figure out which zero to use
first, in this case x=4 is good.
Example 2 cont.
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Using synthetic division
4
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2
2
-7
-10
24
8
4
-24
1
-6
0
F(x)=(x-4)(2x²+x-6)
Factor 2x²+x-6
o (2x-3)(x+2)
So f(x)=(x-4)(2x-3)(x+2)
Zeros= 4, 3/2, -2
Example 2 cont.
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Sign Chart
Sign
Change
-
-2
Sign
Change
+
3/2
Sign
Change
-
4
+
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You can find the where it is negative or positive from its
end behavior
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Since we wanted to find out when it is greater than 0,
the solutions are (-2,3/2) and (4,∞)
Example 3 Solving Graphically
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Solve x3-6x2 ≤ 2-8x graphically
Rewrite the inequality so it is less than or equal to 0
o x3-6x2+8x-2 ≤ 0
Type in x3-6x2+8x-2 into the y1 of the graph on your calculator
o Zeros are approximately 0.32, 1.46, and 4.21
Since we want when it is less than 0, we want all of the numbers below the
x-axis on the graph
o Solution: (-∞,0.32]
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and [1.46, 4.21]
Remember, use hard brackets because those points are solutions too!
Example 4 Solving with Unusual
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The inequalities associated with a strictly positive polynomial function such
as f(x) = (x2+7)(2x2+1) have strange solutions
o (x2+7)(2x2+1) > 0 is all real numbers
o (x2+7)(2x2+1) ≥ 0 is all real numbers
o (x2+7)(2x2+1) < is no solution
o (x2+7)(2x2+1) ≤ is no solution
Example 4 Cont.
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The inequalities associated with a nonnegative polynomil function such as
o (x2-3x+3)(2x+5) > 0 is (-∞,-5/2) and (-5/2,∞)
o (x2-3x+3)(2x+5) ≥ 0 is all real numbers
o (x2-3x+3)(2x+5) < 0 has no solution
o (x2-3x+3)(2x+5) ≤ 0 is a single number, -5/2
Example 5 Creating Sign Charts
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Let f(x) = (2x+1)/((x+3)(x-1)). Find when the function is (a) zero (b)
undefined. Then make a sign chart to find when it is positive or negative.
(a). Real zeros of the function are the real zeros of the numerator. in this case 2
x+1 is the numerator
(b). f(x) is undefined when the denominator is 0. Since (x+3)(x-1) is the
denominator, it is undefined at x = -3 or x = 1.
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Sign Chart
Potential
Sign Change
Potential
Sign Change
Potential
Sign Change
-3
-1/2
1
Example 5 cont.
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Sign chart with undefined, zeros, positive, and negative
(-)
(-)(-)
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(-)
und.
(+)(-)
-3
+
(+)
0
-1/2
(+)(-)
-
(+)
und.
1
(+)(+)
+
f(x) is negative if x < 3 or -1/2 < x < 1, so the solutions are (-∞, -3) and (1/2, 1)
f(x) is positive if -3 < x < -1/2 or x > 1, so the solutions are (-3, -1/2) and
(1,∞)
Example 6 Solve by Combining Fractions
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Solve (5/(x+3))+(3/(x-1)) < 0
5
3
x+3
+
5(x-1)
(x+3)(x-1)
x-1
<
0
3(x+3)
+
(x+3)(x-1)
5(x-1) + 3(x+3)
(x+3)(x-1)
<0
Original Inequality
<0
Use LCD to rewrite fractions
Example 6 cont.
5x-5+3x+9
(x+3)(x-1)
<0
Distributive property
8x+4
(x+3)(x-1)
<0
Simplify
<0
Divide both sides by 4
2x+1
(x+3)(x-1)
Solution: (-∞, -3) and (-1/2, 1).
Example 7 Inequalities Involving
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Solve (x-3)√(x+1) ≥ 0.
Because of the factor √(x+1), f(x) is undefined if x < -1.
The zeros of f are 3 and -1.
Sign Chart:
0
Undefined
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-1
(-)(+)
Negative
Solution: {-1} and [3, ∞)
0
3
(+)(+)
Positive
Example 8 Inequalities with Absolute
Value
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Solve x-2
x+3
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≤0
Because x+3 is in the denominator, f(x) is undefined if x
= -3.
The only zeros of f is 2.
(-)
Negative
(-)
+
Negative
(+)
+
Positive
• Solution: (-∞, -3) and (-3,2]
Matching Game
The link for the game can be found here
http://quizlet.com/18669267/scatter/
A = 60 seconds or less
B = in between 60.1 and 90 seconds
C = in between 90.1 and 120 seconds
D = in between 120.1 and 150 seconds
F = Anything greater than 150.1 seconds
Quiz
x2+5/x
A.) (x + 5)/x3
B.) (x3 + 5)/x
C.) (x + 5)3/x
2.) Which one of these is a possible rational zero of the polynomial.
2x3+x2-4x-3
A.) ±4
B.) ±2
C.) ±3
D.) All the above
3.) Determine the x values that cause the polynomial function to be a zero.
f(x) = (2x2+5)(x-8)2(x+1)3
A.) 8
B.) -1
C.) 5
D.) A and B
E.) All the above
Quiz Page 2
4.) The graph of f(x) = x4(x+3)2(x-1)3 changes sign at x = 0.
A.) True
B.) False
5.) Which is a solution to x2 < x
A.) (1, ∞)
B.) (0,1)
C.) (0, ∞)
6.) Solve the inequality. x|x - 2| > 0
A.) (0,2)U(2,∞)
B.) (-∞, 2)U(2,∞)
7.) Solve the polynomial inequality. x3 - x2 - 2x ≥ 0
A.) [-2,0]U[1,∞)
B.) [-1,0]U[2,∞)
C.) [0,1]U[2,∞)
Quiz Page 3
8.) Complete the factoring if needed and solve the polynomial inequality.
(x + 1)(x2 - 3x + 2) < 0
A.) [-1,0]U[2,∞)
B.) (-∞,0)U(2,3)
C.) (-∞,-1)U(1,2)
9.) Dunder Mifflin Paper Company wishes to design paper boxes with a volume of not more than
100 in3. Squares are to be cut from the corners of a 12-in. by 15-in. piece of cardboard, with the
flaps folded up to make an open box. What size squares should be cut from the cardboard.
A.) 0 in. ≤ x ≤ 0.69 in.
B.) 0 in. ≥ x ≥ 0.69 in.
C.) 4.20 ≤ x ≤ 6 in.
D.) 4.20 ≥ x ≥ 6 in.
E.) A and C
F.) B and D
10.) Solve the polynomial inequality. 2x3 - 5x2 - x + 6 > 0
A.) (-1, 3/2)U(2,∞)
B.) [-1, 3/2]U[2,∞]
C.) (-1, 3/2]U[2,∞)
1.) B
2.) C 3.) D 4.) False 5.) B 6.) A 7.) B 8.) C 9.) E 10.)
Work Cited
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Precalculus Graphical, Numerical, Algebraic; Eighth Edition