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The function ๐‘“(๐‘ฅ)=โˆ’3๐‘ฅ4+16๐‘ฅ3+24๐‘ฅ2โˆ’192๐‘ฅ+10 has critical points at some of the following ๐‘ฅ -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 ๐‘ฅ=โˆ’6 is Answer 1 Question 3 The point ๐‘ฅ=โˆ’2 is Answer 2 Question 3 The point ๐‘ฅ=0 is Answer 3 Question 3 The point ๐‘ฅ=2 is Answer 4 Question 3 The point ๐‘ฅ=4 is Answer 5 Question 3

Question

The function

๐‘“(๐‘ฅ)=โˆ’3๐‘ฅ4+16๐‘ฅ3+24๐‘ฅ2โˆ’192๐‘ฅ+10

has critical points at some of the following ๐‘ฅ -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 ๐‘ฅ=โˆ’6 is Answer 1 Question 3

The point ๐‘ฅ=โˆ’2 is Answer 2 Question 3

The point ๐‘ฅ=0 is Answer 3 Question 3

The point ๐‘ฅ=2 is Answer 4 Question 3

The point ๐‘ฅ=4 is Answer 5 Question 3

...expand
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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‘“(๐‘ฅ)=โˆ’3๐‘ฅ^4+16๐‘ฅ^3+24๐‘ฅ^2โˆ’192๐‘ฅ+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:

  1. ๐‘ฅ=โˆ’6: Substituting x = -6 into the derivative equation gives a non-zero value, so x = -6 is not a critical point.

  2. ๐‘ฅ=โˆ’2: Substituting x = -2 into the derivative equation gives a non-zero value, so x = -2 is not a critical point.

  3. ๐‘ฅ=0: Substituting x = 0 into the derivative equation gives a non-zero value, so x = 0 is not a critical point.

  4. ๐‘ฅ=2: Substituting x = 2 into the derivative equation gives a non-zero value, so x = 2 is not a critical point.

  5. ๐‘ฅ=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.

This problem has been solved

Similar Questions

The function ๐‘“(๐‘ฅ)=โˆ’3๐‘ฅ4+16๐‘ฅ3+24๐‘ฅ2โˆ’192๐‘ฅ+10 has critical points at some of the following ๐‘ฅ -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 ๐‘ฅ=โˆ’6 is Answer 1 Question 3 The point ๐‘ฅ=โˆ’2 is Answer 2 Question 3 The point ๐‘ฅ=0 is Answer 3 Question 3 The point ๐‘ฅ=2 is Answer 4 Question 3 The point ๐‘ฅ=4 is Answer 5 Question 3

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