Good afternoon,
I need a bit of help.
So ive got my first derivative which is
6x^2-6x-12
and the stationary points ive found are x=-1 and x=2 at which points the value of the derivative is zero.
So im attempting to classify the points. From my derivative tests
the value of -2 should give a value of 24 which is positive.
6(-2)^2-6(-2)-12 = 24
and a value of 3 should give 24 also which is again positive.
6(3)^2-6(3)-12 = 24
So I would classify my stationary points as a point of inflection?
Is that correct or did I need to simplify the derivative further before applying this test?
And have I gone wrong at any point along the way that you can see?
You say dy/dx = 6(x^2 - x - 2)
so the function is horizontal when
(x-2)(x+1) = 0
or
x = 2 and x = -1 Agreed
NOW
you can test it like you did, but usually what you do is take the second derivative. If it is +, the function is at the bottom and about to start up. If it is negative, the function is at a hump and about to head down. If it is zero, then inflection point.
so (I am ignoring that 6 we factored out)
d^2y/dx^2 = 2 x - 1
at x = -1 this is
NEGATIVE, so function at a maximum
at x = 2, this is POSITIVE so function is at a minimum and about to head up.
Good afternoon!
Based on the information you've provided, let's go through the steps to classify the stationary points correctly.
Step 1: Find the first derivative
You correctly found the first derivative of the function:
f'(x) = 6x^2 - 6x - 12
Step 2: Find the critical points
To find the critical points, we need to set the derivative equal to zero and solve for x:
6x^2 - 6x - 12 = 0
You correctly found the critical points to be x = -1 and x = 2.
Step 3: Use the second derivative test
To classify the stationary points, we can use the second derivative test. The second derivative, f''(x), is obtained by taking the derivative of f'(x), which is the first derivative.
f''(x) = 12x - 6
Now let's substitute the critical points into the second derivative to determine the nature of the stationary points.
For x = -1:
f''(-1) = 12(-1) - 6 = -12 - 6 = -18 (negative)
For x = 2:
f''(2) = 12(2) - 6 = 24 - 6 = 18 (positive)
Step 4: Analyzing the results
The second derivative test states that if f''(x) > 0, the function is concave up, which implies a minimum. If f''(x) < 0, the function is concave down, which implies a maximum. If the second derivative is zero, the test is inconclusive.
In your case, f''(-1) = -18 (negative), indicating a concave down shape, suggesting a local maximum at x = -1. And f''(2) = 18 (positive), implying a concave up shape, which suggests a local minimum at x = 2.
Therefore, your classification of the points is correct. The point x = -1 is a local maximum, and the point x = 2 is a local minimum.
Regarding your question about simplifying the derivative further before applying the second derivative test, it is not necessary. The derivative you have is already simplified and can be used directly to find the critical points and classify them.
It seems like you have followed the correct steps and performed the calculations accurately. Well done! If you have any further questions, feel free to ask.