Consider the function f(x) = -2 x^3 + 36 x^2 - 162 x + 2. There are two critical points, A and B where A < B:
A = and B =
f''(A)=
f''(B)=
Therefore f(x) has a relative at A (MAX or MIN)
and a relative at B (MAX or MIN).
plain old power rule here.
f'(x) = -6x^2 + 72x - 162
= -6(x^2 - 12x + 27)
f"(x) = -12x + 72
= -12(x-6)
So, f' = 0 when x = 3 or 9
f"(3) > 0 so f is concave up (min)
f"(9) < 0 so f is concave down (max)
Note that you can also just rely on what you know about the general shape of cubics to identify the max/min-ness of the extrema.
To find the critical points for the function f(x) = -2x^3 + 36x^2 - 162x + 2, we need to take the derivative and set it equal to zero.
First, let's find the first derivative of f(x):
f'(x) = -6x^2 + 72x - 162
Now, let's set f'(x) equal to zero and solve for x:
-6x^2 + 72x - 162 = 0
To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula. However, in this case, factoring the equation is a bit challenging, so we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values for a, b, and c:
a = -6, b = 72, c = -162
We get:
x = (-72 ± √(72^2 - 4(-6)(-162))) / 2(-6)
x = (-72 ± √(5184 - 1944)) / -12
x = (-72 ± √3240) / -12
x = (-72 ± 9√10) / -12
Simplifying further:
x = (6 ± √10) / 2
So, we have two critical points:
A = (6 - √10) / 2 ≈ 1.172
B = (6 + √10) / 2 ≈ 4.828
To find the second derivative of f(x), we need to take the derivative of f'(x):
f''(x) = -12x + 72
Now, let's evaluate f''(A) and f''(B):
f''(A) = -12(A) + 72
= -12((6 - √10) / 2) + 72
= -12(6/2 - √10/2) + 72
= -12(3 - √10/2) + 72
= -36 + 6√10 + 72
= 6√10 + 36
f''(B) = -12(B) + 72
= -12((6 + √10) / 2) + 72
= -12(6/2 + √10/2) + 72
= -12(3 + √10/2) + 72
= -36 - 6√10 + 72
= -6√10 + 36
Next, let's determine the nature of the critical points A and B.
Since f''(A) is positive (6√10 + 36 > 0), the second derivative test tells us that at the critical point A, f(x) has a relative minimum.
Similarly, since f''(B) is negative (-6√10 + 36 < 0), the second derivative test tells us that at the critical point B, f(x) has a relative maximum.
Therefore, we can conclude that f(x) has a relative minimum at A and a relative maximum at B.