Question

Find the relative extrema of the function and classify each as a maximum or minimum. f(x) = x3 - 12x - 1 Relative maximum: (5, 64); relative minimum: (2, -17) O Relative maximum: (5, 64); relative minimum: (-3, 8) Relative minimum: (-2, 15); relative maximum: (2, -17) Relative maximum: (-2, 15); relative minimum: (2, -17)

Answer 1

The function f(x) = x^3 - 12x - 1 has a relative maximum at the point (-2, 15) and a relative minimum at the point (2, -17). These points are found by setting the first derivative to zero and then using the second derivative to determine the concavity at these points.

Step-by-step explanation:

To find the relative extrema of the function f(x) = x^3 - 12x - 1, we need to calculate its first and second derivatives. The first derivative f'(x) gives us the slope of the tangent line to the function at any point x and is used to determine where the slope is zero, indicating possible relative extrema. The second derivative f''(x) is used to determine the concavity of the function, which helps us classify the extrema as maximums or minimums.

Firstly, we find the first derivative f'(x) = 3x^2 - 12. Setting this derivative equal to zero gives us the potential points of relative extrema: 3x^2 - 12 = 0 which simplifies to x^2 = 4. The solutions to this equation are x = 2 and x = -2.

Secondly, we calculate the second derivative f''(x) = 6x. We evaluate this second derivative at the critical points x = 2 and x = -2. For x = 2, the second derivative f''(2) = 6(2) = 12 which is positive, indicating a relative minimum. For x = -2, we get f''(-2) = 6(-2) = -12 which is negative, indicating a relative maximum.

Now we plug these x values into the original function to get their corresponding y values. For x = 2, f(2) = 2^3 - 12(2) - 1 = -17, and for x = -2, f(-2) = (-2)^3 - 12(-2) - 1 = 15. Therefore, the relative maximum is (-2, 15) and the relative minimum is (2, -17).

Answer 2

To find the relative minimum and maximum, we need to find the derivative for the equation of the function:

`f(x)=x^3-12x-1`

Then, the derivative is (following the rules of derivatives):

`3x^2-12=3(x^2-4)`

To find the relative maximum and minimum, we need to equate the derivative to zero, and solve for x:

`3(x^2-4)=0\Rightarrow x^2-4=0\Rightarrow x^2=4\Rightarrow x=\pm2`

In this case, we have two values for x, x = -2, and x = 2.

Substituting these values in the original equation, we have:

`f(-2)=3\cdot(-2)^2-12(-2)-1\Rightarrow f(-2)=15`

And

`f(2)=3\cdot(2)^2-12\cdot(2)-1\Rightarrow f(2)=-17`

In this case, we have two pair of values:

(-2, 15), and (2, -17).

If we obtain the second derivative for the original value, we have that is 6x.

If we substitute x = -2 in the second derivative, we have 6 * (-2) = -12. Since the value is negative, the point (-2, 15) is a Relative Maximum.

Likewise, if we make x = 2, we have 6 * (2) = 12. Since this value is positive, we have a relative minimum (2, -17).

Therefore:

Relative maximum: (-2, 15).

Relative minimum: (2, -17).

So the correct option is the last one.