the function defined by f(x)=x^3 - 3x^2 for all real numbers has a relative maximum at x = ?
To find the relative maximum of a function, we need to find the critical points by taking the derivative of the function and setting it equal to zero.
Given:
f(x) = x^3 - 3x^2
Taking the derivative of f(x) with respect to x:
f'(x) = 3x^2 - 6x
Now we set f'(x) equal to zero and solve for x:
3x^2 - 6x = 0
Factor out 3x:
3x(x - 2) = 0
Setting each factor equal to zero:
3x = 0 or x - 2 = 0
x = 0 or x = 2
These are the critical points of the function.
To determine if each critical point represents a relative maximum or minimum, we can analyze the sign of the second derivative. If the second derivative is positive, it represents a relative minimum. If the second derivative is negative, it represents a relative maximum. If the second derivative is zero, we need to consider higher-order derivatives or use other methods.
Taking the second derivative of f(x):
f''(x) = 6x - 6
Now we substitute the critical points into f''(x):
For x = 0:
f''(0) = 6(0) - 6 = -6
For x = 2:
f''(2) = 6(2) - 6 = 6
Since f''(0) = -6 is negative and f''(2) = 6 is positive, we can conclude that x = 0 is a relative maximum and x = 2 is a relative minimum.
Therefore, the relative maximum of the function f(x) = x^3 - 3x^2 occurs at x = 0.
To find the relative maximum of a function, we need to determine the critical points. In this case, the critical points occur when the derivative of the function equals zero or is undefined. Let's find the derivative of the function f(x) = x^3 - 3x^2:
f'(x) = 3x^2 - 6x
We set f'(x) = 0 to find the critical points:
3x^2 - 6x = 0
Factoring out 3x, we get:
3x(x - 2) = 0
Setting each factor equal to zero, we find:
3x = 0 or x - 2 = 0
Solving for x, we get:
x = 0 or x = 2
Now, we need to check the second derivative to determine whether the critical points are relative maxima or minima. Let's find f''(x):
f''(x) = (3x^2 - 6x)' = 6x - 6
Now substituting the critical points:
f''(0) = 6(0) - 6 = -6
f''(2) = 6(2) - 6 = 6
Since f''(0) < 0 and f''(2) > 0, we can conclude that x = 0 is a relative maximum and x = 2 is a relative minimum.
Therefore, the function f(x) = x^3 - 3x^2 has a relative maximum at x = 0.