I need an expiation on a math concept, because I know the right answer, I just don't know why it is
The problem says to find the absolute maximum and minimum of the function f(x)=x^3-12x over the interval [-5, 2]
I found the 1st der. 3x^2-12 and set equal to zero, got the x values -2 and 2
f(-2)= 16 f(2)= -16 f(-5)=-65
The second der. is 6x
6(-2)=-12
6(2)= 12
6(-5)= -30
I read somewhere that if f''(x) > 0 it was a minimum, and if f''(x) < 0 it was a maximum. So by that, I would say that the max was x= -5, at -65
and min was x=-2, at 16
However, the max was said to be what I thought the min was and vice versa, but why? I know for other parts of my homework my mins and maxs were correct, so why is it not so for this problem?
Thanks
The relative max/min occur at x = -2,2 as you said.
Since f"(-2) < 0, f(x) has a max at x = -2 (f is concave down)
Since f"(2) > 0, f(x) has a minimum at x=2 (f is concave up)
f(-5) = -65, so the absolute minimum of f(x) on [-5,2] is at x = -5, but that is just because f(-5) < f(2), not because is a critical point. f'(-5)≠0
See the graph at
http://www.wolframalpha.com/input/?i=x^3-12x
To find the absolute maximum and minimum of a given function over a specified interval, we need to examine the critical points (where the derivative is zero or does not exist) and the endpoints of the interval.
In your solution, you correctly found the critical points of the function f(x) = x^3 - 12x by setting the derivative equal to zero: 3x^2 - 12 = 0. Solving this equation, you found x = -2 and x = 2.
Now, to determine whether these critical points are relative maximums or minimums, we can use the second derivative test. The second derivative, f''(x), gives us information about the concavity of the function.
When f''(x) > 0, the function is concave up, and this indicates a relative minimum. When f''(x) < 0, the function is concave down, which indicates a relative maximum.
In your work, you correctly found the second derivative of the function f(x) = x^3 - 12x to be f''(x) = 6x.
Plugging in the values of the critical points into f''(x), we have:
f''(-2) = 6(-2) = -12
f''(2) = 6(2) = 12
f''(-2) is negative, meaning the function is concave down at x = -2. So, this is a relative maximum.
f''(2) is positive, indicating that the function is concave up at x = 2. Hence, this is a relative minimum.
Now we need to consider the endpoints of the interval, [-5, 2].
At x = -5, we have f(-5) = -65.
At x = 2, we have f(2) = -16.
Comparing these four values, we can now determine the absolute maximum and minimum:
The absolute maximum is the largest value among the function values at the critical points and the endpoints. In this case, the largest value is f(-5) = -65. So, the absolute maximum occurs at x = -5, and the maximum value is -65.
The absolute minimum is the smallest value among the function values at the critical points and the endpoints. In this case, the smallest value is f(2) = -16. Therefore, the absolute minimum occurs at x = 2, and the minimum value is -16.
Therefore, the maximum value at -65 occurs at x = -5, and the minimum value at -16 occurs at x = 2.
If your answer differed from this, it is likely due to an error in interpreting the concavity or comparing the function values correctly. Double-checking your calculations and comparisons may help you identify where the mistake occurred.