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y is the smallest positive integer such that for any integer x > y, the quantity x3 - 11x2 + 39x - 45 is positive. What is the value of y?
Correct answer is '5'. Can you explain this answer?
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Verified Answer
y is the smallest positive integer such that for any integer x > y...
Consider the given expression,
x3 - 11x2 + 39x- 45
For x = 3, the expression reduces to zero
∴ (x - 3) is a factor.
Now, the expression can be written as, (x-3)2(x - 5)
Since, (x - 3)2 is always positive.
∴ The expression is positive only when x > 5.
∴ x = 6 and y = 5
Answer: 5
x3 - 11x2 + 39x- 45
For x = 3, the expression reduces to zero
∴ (x - 3) is a factor.
Now, the expression can be written as, (x-3)2(x - 5)
Since, (x - 3)2 is always positive.
∴ The expression is positive only when x > 5.
∴ x = 6 and y = 5
Answer: 5
Most Upvoted Answer
y is the smallest positive integer such that for any integer x > y...
The given expression is x^3 - 11x^2 + 39x - 45.
To find the smallest positive integer y such that the expression is always positive, we need to analyze the behavior of the expression for different values of x.
1. Analyzing the Expression
We can start by factoring the expression to gain some insights:
x^3 - 11x^2 + 39x - 45 = (x - 5)(x^2 - 6x + 9) + (4x + 0)
From this factorization, we observe that the expression can be written as the sum of two terms: (x - 5)(x^2 - 6x + 9) and 4x.
2. Analyzing the First Term
The first term, (x - 5)(x^2 - 6x + 9), can be further factored as (x - 5)(x - 3)(x - 3).
From this factorization, we can see that the first term is equal to zero at x = 5 and x = 3. These values are called the critical points.
3. Analyzing the Second Term
The second term, 4x, is a linear expression that increases with x. It is always positive for positive values of x.
4. Analyzing the Expression as a Whole
Combining the above analysis, we can make the following observations:
- For x < 3,="" both="" the="" first="" term="" and="" the="" second="" term="" are="" negative,="" resulting="" in="" a="" negative="" />
- For 3 < x="" />< 5,="" the="" first="" term="" is="" positive="" (since="" it="" is="" the="" product="" of="" three="" negative="" factors)="" and="" the="" second="" term="" is="" positive,="" resulting="" in="" a="" positive="" />
- For x > 5, the first term is positive and the second term is positive, resulting in a positive expression.
5. Concluding the Smallest Positive Integer
From the above analysis, we can conclude that for any value of x greater than 5, the expression x^3 - 11x^2 + 39x - 45 is positive. Therefore, the smallest positive integer y such that the expression is always positive is 5.
Summary:
- The given expression x^3 - 11x^2 + 39x - 45 can be factored as (x - 5)(x^2 - 6x + 9) + (4x + 0).
- The first term (x - 5)(x^2 - 6x + 9) has critical points at x = 5 and x = 3.
- The second term 4x is always positive for positive values of x.
- Combining these observations, we find that the expression is positive for x > 5.
- Therefore, the smallest positive integer y such that the expression is always positive is 5.
To find the smallest positive integer y such that the expression is always positive, we need to analyze the behavior of the expression for different values of x.
1. Analyzing the Expression
We can start by factoring the expression to gain some insights:
x^3 - 11x^2 + 39x - 45 = (x - 5)(x^2 - 6x + 9) + (4x + 0)
From this factorization, we observe that the expression can be written as the sum of two terms: (x - 5)(x^2 - 6x + 9) and 4x.
2. Analyzing the First Term
The first term, (x - 5)(x^2 - 6x + 9), can be further factored as (x - 5)(x - 3)(x - 3).
From this factorization, we can see that the first term is equal to zero at x = 5 and x = 3. These values are called the critical points.
3. Analyzing the Second Term
The second term, 4x, is a linear expression that increases with x. It is always positive for positive values of x.
4. Analyzing the Expression as a Whole
Combining the above analysis, we can make the following observations:
- For x < 3,="" both="" the="" first="" term="" and="" the="" second="" term="" are="" negative,="" resulting="" in="" a="" negative="" />
- For 3 < x="" />< 5,="" the="" first="" term="" is="" positive="" (since="" it="" is="" the="" product="" of="" three="" negative="" factors)="" and="" the="" second="" term="" is="" positive,="" resulting="" in="" a="" positive="" />
- For x > 5, the first term is positive and the second term is positive, resulting in a positive expression.
5. Concluding the Smallest Positive Integer
From the above analysis, we can conclude that for any value of x greater than 5, the expression x^3 - 11x^2 + 39x - 45 is positive. Therefore, the smallest positive integer y such that the expression is always positive is 5.
Summary:
- The given expression x^3 - 11x^2 + 39x - 45 can be factored as (x - 5)(x^2 - 6x + 9) + (4x + 0).
- The first term (x - 5)(x^2 - 6x + 9) has critical points at x = 5 and x = 3.
- The second term 4x is always positive for positive values of x.
- Combining these observations, we find that the expression is positive for x > 5.
- Therefore, the smallest positive integer y such that the expression is always positive is 5.
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y is the smallest positive integer such that for any integer x > y, the quantity x3 - 11x2 + 39x - 45 is positive. What is the value of y?Correct answer is '5'. Can you explain this answer?
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