Which points are the best approximations of the relative maximum and minimum of the function?
F(x)= x^3+6x^2-36x
To find the relative maximum and minimum points of the function F(x) = x^3 + 6x^2 - 36x, we need to find the critical points and then evaluate the function at those points.
First, let's find the derivative of F(x):
F'(x) = 3x^2 + 12x - 36
Next, we set F'(x) equal to zero and solve for x to find the critical points:
3x^2 + 12x - 36 = 0
Factorizing the equation, we get:
3(x^2 + 4x - 12) = 0
Setting each factor equal to zero, we have:
x^2 + 4x - 12 = 0
Using the quadratic formula, we can solve for x:
x = (-4 ± √(4^2 - 4(1)(-12))) / (2(1))
x = (-4 ± √(16 + 48)) / 2
x = (-4 ± √64) / 2
x = (-4 ± 8) / 2
Simplifying further, we get:
x = (-4 + 8) / 2 = 4/2 = 2
x = (-4 - 8) / 2 = -12/2 = -6
So the critical points are x = 2 and x = -6.
Now, let's evaluate F(x) at these critical points to find the corresponding y-values:
F(2) = 2^3 + 6(2)^2 - 36(2)
= 8 + 24 - 72
= -40
F(-6) = (-6)^3 + 6(-6)^2 - 36(-6)
= -216 + 216 + 216
= 0
Therefore, the point (2, -40) is the approximate relative minimum point, and the point (-6, 0) is the approximate relative maximum point for the function F(x) = x^3 + 6x^2 - 36x.