in 1-x-x^2, it gets differentiated to -1-2x, then 0= -1-2x, and after some dividing you get -1/2,-3/4 (correct me if I was incorrect), but how do you tell if they are the maxima or minima?
-
x=-1/2 , y = 5/4
it's maximum because the parabola y = -x^2 +...
it's maximum because the parabola y = -x^2 +...
-
What you found are the critical numbers. What you need to do now is determine what you want to find in your interval.
If you want to find absolute maxima and absolute minima, then plug in your critical numbers as well as the endpoints of your interval and plug into the original equation (not the derivative) Then just compare all 4 points, and see which one is the highest and which is the lowest. The highest being the maxima and the lowest being the minima
If you want to find relative extrema, all you have to do is take your critical numbers, and set intervals based on them. In your problem, your intervals will be (neg infinity < x < -3/4) (-3/4 < x < -1/2) (-1/2 < x < pos infinity)
Now choose 3 test values based on your intervals. (-1, -0.6, 1)
Then plug in your 3 test values into your DERIVATIVE. If the answer is positive, you have a maxima, and if it is negative, you have a minima.
If you want to find absolute maxima and absolute minima, then plug in your critical numbers as well as the endpoints of your interval and plug into the original equation (not the derivative) Then just compare all 4 points, and see which one is the highest and which is the lowest. The highest being the maxima and the lowest being the minima
If you want to find relative extrema, all you have to do is take your critical numbers, and set intervals based on them. In your problem, your intervals will be (neg infinity < x < -3/4) (-3/4 < x < -1/2) (-1/2 < x < pos infinity)
Now choose 3 test values based on your intervals. (-1, -0.6, 1)
Then plug in your 3 test values into your DERIVATIVE. If the answer is positive, you have a maxima, and if it is negative, you have a minima.
-
I dont know how you got -3/4
0 = - 1 - 2x
2x = - 1
x = -1/2
look at what the first derivative does when x < -1/2
You can try and put in nice values to test it such as x = -1
-1 - 2x = -1 -2(- 1) = -1 + 2 = 1
The first derivative > 0, so the function is increasing from (- infinity, -1/2)
Try value for x > -1/2 such as x = 0
-1 - 2(0) = -1 - 0 = -1
The first derivative < 0, so the function is decreasing from (-1/2, infinity)
The function increases to x = 1/2 and then decreases
x = - 1/2 is an absolute maxima
0 = - 1 - 2x
2x = - 1
x = -1/2
look at what the first derivative does when x < -1/2
You can try and put in nice values to test it such as x = -1
-1 - 2x = -1 -2(- 1) = -1 + 2 = 1
The first derivative > 0, so the function is increasing from (- infinity, -1/2)
Try value for x > -1/2 such as x = 0
-1 - 2(0) = -1 - 0 = -1
The first derivative < 0, so the function is decreasing from (-1/2, infinity)
The function increases to x = 1/2 and then decreases
x = - 1/2 is an absolute maxima
-
Take the second differential of the equation, and input (if any) x values into it that you have previously calculated via the first differential. If this value is negative, it is a maxima, and vice versa. We know this because f'''(x) shows the rate of change of the gradient. If the gradient is changing to become negative, then by a synonymous statement, the change of gradient is negative and vice versa.
-
Look at the leading coefficient in the highest degree. If it is positive, the graph opens upwards and the answer is the minima. If it is negative, the graph opens downwards, and the number is the maxima. The leading coefficient is -1 in the -x^2. This is negative and -1/2 is the maxima.