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consider y=2x/1+x², where x is real , then the range of expression y²+y-2 is [a,b]. Find b - 4a. Please solve this quadratic problem
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consider y=2x/1+x², where x is real , then the range of expression y²+...
So b= 0 and a = -9/4 and hence b - 4a = 9
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consider y=2x/1+x², where x is real , then the range of expression y²+...
Problem Statement: Consider y=2x/1 x², where x is real, then the range of expression y² y-2 is [a,b]. Find b - 4a.
Solution:
Step 1: Find the range of y
y=2x/1 x² = 2/x
Since x is real, we know that x can be positive, negative or zero. However, we cannot have y=0 as it would make the denominator of the expression undefined.
Therefore, the range of y is (-∞, 0) U (0, ∞)
Step 2: Find the range of y² y-2
Substituting y=2/x in y² y-2, we get:
y² y-2 = (2/x)² (2/x) - 2
= 4/x² (2/x) - 2
= 8/x³ - 2
Since x can be positive, negative or zero, we need to consider each case separately.
Case 1: x > 0
In this case, the range of y is (0, ∞)
Substituting x=1, we get:
y² y-2 = 8/1³ - 2 = 6
Substituting x→∞, we get:
y² y-2 → 0
Therefore, the range of y² y-2 for x > 0 is (0, 6]
Case 2: x < />
In this case, the range of y is (-∞, 0)
Substituting x=-1, we get:
y² y-2 = 8/(-1)³ - 2 = -6
Substituting x→-∞, we get:
y² y-2 → 0
Therefore, the range of y² y-2 for x < 0="" is="" [-6,="" />
Case 3: x = 0
In this case, y is undefined and y² y-2 is also undefined.
Step 3: Find b - 4a
Therefore, the range of y² y-2 is [-6, 6]
b - 4a = 6 - 4(-6) = 30
Therefore, the value of b - 4a is 30.
Solution:
Step 1: Find the range of y
y=2x/1 x² = 2/x
Since x is real, we know that x can be positive, negative or zero. However, we cannot have y=0 as it would make the denominator of the expression undefined.
Therefore, the range of y is (-∞, 0) U (0, ∞)
Step 2: Find the range of y² y-2
Substituting y=2/x in y² y-2, we get:
y² y-2 = (2/x)² (2/x) - 2
= 4/x² (2/x) - 2
= 8/x³ - 2
Since x can be positive, negative or zero, we need to consider each case separately.
Case 1: x > 0
In this case, the range of y is (0, ∞)
Substituting x=1, we get:
y² y-2 = 8/1³ - 2 = 6
Substituting x→∞, we get:
y² y-2 → 0
Therefore, the range of y² y-2 for x > 0 is (0, 6]
Case 2: x < />
In this case, the range of y is (-∞, 0)
Substituting x=-1, we get:
y² y-2 = 8/(-1)³ - 2 = -6
Substituting x→-∞, we get:
y² y-2 → 0
Therefore, the range of y² y-2 for x < 0="" is="" [-6,="" />
Case 3: x = 0
In this case, y is undefined and y² y-2 is also undefined.
Step 3: Find b - 4a
Therefore, the range of y² y-2 is [-6, 6]
b - 4a = 6 - 4(-6) = 30
Therefore, the value of b - 4a is 30.
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consider y=2x/1+x², where x is real , then the range of expression y²+y-2 is [a,b]. Find b - 4a. Please solve this quadratic problem
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