Find the relative extrema, if any, of the function. Use the second derivative test, if applicable. (If an answer does not exist, enter DNE.)
g(x)=x^3-6x
relative maximum (x, y) = ( , )
relative minimum (x, y) = ( , )
To find the relative extrema of the function g(x) = x^3 - 6x, we first need to find the critical points of the function. The critical points occur when the derivative of the function is either zero or does not exist.
Step 1: Find the derivative of g(x)
g'(x) = 3x^2 - 6
Step 2: Set g'(x) = 0 and solve for x to find the critical points.
3x^2 - 6 = 0
3x^2 = 6
x^2 = 2
x = ±√2
So, the critical points are x = √2 and x = -√2.
Step 3: Determine the nature of the critical points using the second derivative test.
To apply the second derivative test, we need to find the second derivative of g(x).
g''(x) = (d/dx)(3x^2 - 6) = 6x
Now, substitute the critical points into g''(x) to determine the nature of the critical points.
For x = √2:
g''(√2) = 6(√2) = 6√2 > 0
For x = -√2:
g''(-√2) = 6(-√2) = -6√2 < 0
According to the second derivative test, when the second derivative is positive, it indicates a relative minimum, and when the second derivative is negative, it indicates a relative maximum.
So, by analyzing the nature of the critical points:
- At x = √2, g(x) has a relative minimum.
- At x = -√2, g(x) has a relative maximum.
Now we can determine the coordinates of the relative extrema:
For the relative minimum:
(x,y) = (√2, g(√2)) = (√2, (√2)^3 - 6(√2)) = (√2, 2 - 6√2)
For the relative maximum:
(x, y) = (-√2, g(-√2)) = (-√2, (-√2)^3 - 6(-√2)) = (-√2, -2 + 6√2)
Therefore, the relative extrema of the function g(x) = x^3 - 6x are:
Relative minimum: (x, y) = (√2, 2 - 6√2)
Relative maximum: (x, y) = (-√2, -2 + 6√2)
To find the relative extrema of the function g(x) = x^3 - 6x, we need to determine the critical points of the function and then apply the second derivative test.
Step 1: Find the derivative of g(x)
g'(x) = 3x^2 - 6
Step 2: Set the derivative equal to zero and solve for x to find the critical points
3x^2 - 6 = 0
3x^2 = 6
x^2 = 2
x = ±√2
So, the critical points are x = √2 and x = -√2.
Step 3: Find the second derivative of g(x)
g''(x) = 6x
Step 4: Substitute the critical points into the second derivative to determine the nature of the extrema
For x = √2:
g''(√2) = 6(√2) = 6√2 > 0
Since the second derivative is positive, g(x) has a relative minimum at x = √2.
For x = -√2:
g''(-√2) = 6(-√2) = -6√2 < 0
Since the second derivative is negative, g(x) has a relative maximum at x = -√2.
Therefore, the relative extrema of the function g(x) = x^3 - 6x are:
Relative maximum: (x,y) = (-√2, g(-√2))
Relative minimum: (x,y) = (√2, g(√2))
To find the y-coordinate for each extrema, substitute the x-values into the original function g(x):
g(-√2) = (-√2)^3 - 6(-√2) = -2√2 - 6√2 = -8√2
g(√2) = (√2)^3 - 6(√2) = 2√2 - 6√2 = -4√2
So, the relative extrema are:
Relative maximum: (x,y) = (-√2, -8√2)
Relative minimum: (x,y) = (√2, -4√2)
Final answer:
Relative maximum: (x,y) = (-√2, -8√2)
Relative minimum: (x,y) = (√2, -4√2)