Which points are the best approximation of the relative maximum and minimum of the function?
f(x)=x^3+3x^2-9x-8
To find the relative maximum and minimum points of a function, we need to analyze its derivative and determine where it equals zero.
Let's start by finding the derivative of the function f(x)=x^3+3x^2-9x-8. We can do this by applying the power rule:
f'(x) = 3x^2 + 6x - 9
Next, we solve for when the derivative equals zero to find the critical points:
3x^2 + 6x - 9 = 0
This quadratic equation can be factored as:
(3x - 3)(x + 3) = 0
From here, we get two solutions: x = 1 and x = -3. These are the critical points of the function.
To determine whether these critical points are relative maximum or minimum, we use the second derivative test. We find the second derivative of the function by finding the derivative of the derivative (f''(x)):
f''(x) = 6x + 6
Now, we substitute the critical points (x = 1 and x = -3) into the second derivative.
For x = 1:
f''(1) = 6(1) + 6 = 12
Since the second derivative is positive (greater than zero), the function f(x) has a relative minimum at x = 1.
For x = -3:
f''(-3) = 6(-3) + 6 = -12
Since the second derivative is negative (less than zero), the function f(x) has a relative maximum at x = -3.
Therefore, the points (-3, f(-3)) and (1, f(1)) are the best approximations of the relative maximum and minimum of the function f(x)=x^3+3x^2-9x-8.
To find the relative maximum and minimum of a function, we can start by finding the critical points. These are the points where the derivative of the function is either zero or undefined. The derivative of the function f(x) = x^3 + 3x^2 - 9x - 8 can be found by applying the power rule for differentiation.
First, find the derivative of f(x):
f'(x) = 3x^2 + 6x - 9
Next, set f'(x) equal to zero and solve for x to find the critical points:
3x^2 + 6x - 9 = 0
This equation can be factored as:
3(x^2 + 2x - 3) = 0
(x + 3)(x - 1) = 0
Setting each factor equal to zero gives us two critical points:
x + 3 = 0 → x = -3
x - 1 = 0 → x = 1
To determine if these critical points are relative maxima or minima, we can use the second derivative test.
To find the second derivative, we differentiate f'(x):
f''(x) = 6x + 6
Now, substitute the critical points into the second derivative to evaluate them:
f''(-3) = 6(-3) + 6 = -12
f''(1) = 6(1) + 6 = 12
If f''(x) > 0 at a critical point, the function has a relative minimum at that point. Conversely, if f''(x) < 0, the function has a relative maximum at that point.
Based on the values of f''(-3) = -12 and f''(1) = 12, we can conclude:
- f''(-3) < 0, so the point (-3, f(-3)) = (-3, 10) is a relative maximum.
- f''(1) > 0, so the point (1, f(1)) = (1, -13) is a relative minimum.
Therefore, the best approximations for the relative maximum and minimum of the function f(x) = x^3 + 3x^2 - 9x - 8 are:
Relative maximum: (-3, 10)
Relative minimum: (1, -13)