The function ๐(๐ฅ)=โ3๐ฅ4+16๐ฅ3+24๐ฅ2โ192๐ฅ+10 has critical points at some of the following ๐ฅ -values. In each case identify whether the point is a critical point, and if so find out if it is a maximum, minimum or neither maximum nor minimum. The point ๐ฅ=โ6 is Answer 1 Question 3 The point ๐ฅ=โ2 is Answer 2 Question 3 The point ๐ฅ=0 is Answer 3 Question 3 The point ๐ฅ=2 is Answer 4 Question 3 The point ๐ฅ=4 is Answer 5 Question 3
Question
The function
๐(๐ฅ)=โ3๐ฅ4+16๐ฅ3+24๐ฅ2โ192๐ฅ+10
has critical points at some of the following ๐ฅ -values. In each case identify whether the point is a critical point, and if so find out if it is a maximum, minimum or neither maximum nor minimum.
The point ๐ฅ=โ6 is Answer 1 Question 3
The point ๐ฅ=โ2 is Answer 2 Question 3
The point ๐ฅ=0 is Answer 3 Question 3
The point ๐ฅ=2 is Answer 4 Question 3
The point ๐ฅ=4 is Answer 5 Question 3
Solution
To find the critical points of a function, we first need to find the derivative of the function and set it equal to zero.
The derivative of the function ๐(๐ฅ)=โ3๐ฅ^4+16๐ฅ^3+24๐ฅ^2โ192๐ฅ+10 is:
f'(x) = -12x^3 + 48x^2 + 48x - 192
Setting this equal to zero gives us the critical points:
-12x^3 + 48x^2 + 48x - 192 = 0
This equation can be simplified to:
x^3 - 4x^2 - 4x + 16 = 0
The solutions to this equation are the critical points of the function.
Now, let's check each of the given x-values:
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๐ฅ=โ6: Substituting x = -6 into the derivative equation gives a non-zero value, so x = -6 is not a critical point.
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๐ฅ=โ2: Substituting x = -2 into the derivative equation gives a non-zero value, so x = -2 is not a critical point.
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๐ฅ=0: Substituting x = 0 into the derivative equation gives a non-zero value, so x = 0 is not a critical point.
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๐ฅ=2: Substituting x = 2 into the derivative equation gives a non-zero value, so x = 2 is not a critical point.
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๐ฅ=4: Substituting x = 4 into the derivative equation gives a non-zero value, so x = 4 is not a critical point.
So, none of the given x-values are critical points of the function.
To determine whether a critical point is a maximum, minimum, or neither, we would need to use the second derivative test. However, since none of the given x-values are critical points, we do not need to perform this test.
Similar Questions
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