Decide if the given value of x is a critical number for f, and if so, decide whether the point for x on f is a relative minimum, relative maximum, or neither.
f(x)= 2x^3-3x^2-12x+18; x=2
the ans. is "Critical number, minimum at (2,-2) but don't know how?
f'(x) = 6x^2 - 6x - 12
= 0 for a max/min (critical value)
divide by 6
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
so, yes, x=2 is a critical number, (the other is x = -1)
f '' (x) = 12x-6
f(2) = 2(2^3) - 3(2^2) - 12(2) + 18 = -2
f '' (2) = 24-6 > 0
since at (2,-2) , the value of f '' (2) is positive, the curve is opening upwards at that point and thus
(2,-2) is a minimum point.
http://www.wolframalpha.com/input/?i=plot+y+%3D++2x%5E3-3x%5E2-12x%2B18
I just love this Wolfram website
To determine whether the given value of x is a critical number for f, we need to find the derivative of f(x). The critical numbers occur when the derivative is equal to zero or undefined.
Given: f(x) = 2x^3 - 3x^2 - 12x + 18, x = 2
First, let's find the derivative of f(x) using the power rule:
f'(x) = 6x^2 - 6x - 12
To find the critical numbers, we need to set the derivative equal to zero and solve for x:
6x^2 - 6x - 12 = 0
Next, we can factor the expression:
3(2x^2 - 2x - 4) = 0
Now, set each factor equal to zero:
2x^2 - 2x - 4 = 0
Factoring further:
2(x^2 - x - 2) = 0
(x - 2)(x + 1) = 0
So, x = 2 or x = -1.
Now we can determine the nature of the critical point at x = 2 by considering the concavity of f(x). To do this, we need to find the second derivative of f(x).
Taking the derivative of f'(x) = 6x^2 - 6x - 12, we get:
f''(x) = 12x - 6
Now, substitute x = 2 into f''(x):
f''(2) = 12(2) - 6
f''(2) = 18
Since the second derivative f''(2) is positive, it indicates that f(x) has a relative minimum at x = 2.
Therefore, the given value of x (x = 2) is a critical number for f, and the point (2, -2) represents a relative minimum for the function f(x).
To determine if the given value of x is a critical number for f, we need to check if the derivative of f at that point is equal to zero or undefined.
First, let's find the derivative of f(x):
f'(x) = 6x^2 - 6x - 12
Next, substitute the given value x = 2 into the derivative:
f'(2) = 6(2)^2 - 6(2) - 12
= 24 - 12 - 12
= 0
Since the derivative f'(x) is equal to zero at x = 2, x = 2 is a critical number for f.
To determine whether the critical point (2, -2) is a relative minimum, maximum, or neither, we can use the second derivative test.
Find the second derivative of f(x):
f''(x) = 12x - 6
Evaluate the second derivative at x = 2:
f''(2) = 12(2) - 6
= 24 - 6
= 18
Since f''(2) = 18 is positive, this indicates that the function is concave up at x = 2. According to the second derivative test, if the second derivative is positive, then the critical point is a relative minimum.
Therefore, x = 2 is a critical number for f, and the point (2, -2) is a relative minimum on the function f(x) = 2x^3 - 3x^2 - 12x + 18.