f(x)=3-3x^2 -3 <= x <= 2
How do you find the absolute max and min value?
What does x equal?
Th max and the minimum value will occur where the derivative is zero, OR at one of the endpoints (2 or -3)
When the derivative is zero
df/dx = 0 = -6x
x = 0.
Since this is an upside-down parabola with a maximum value of 3 (at x=0), the minimum value will be where x = -3, which is as far as possible from x = 0
I don't understand how to get the min??
nevermind i got it! thanks!
To find the absolute maximum and minimum values of the given function f(x) = 3 - 3x^2 in the given interval -3 <= x <= 2, you can follow these steps:
Step 1: Find the critical points by taking the derivative of the function f(x) and setting it to zero.
So, let's find f'(x):
f(x) = 3 - 3x^2
f'(x) = 0 - 6x
Setting f'(x) = 0, we get:
0 - 6x = 0
-6x = 0
x = 0
So, the critical point is x = 0.
Step 2: Check the endpoints and the critical points to find the absolute max and min values.
- Evaluate the function at the endpoints of the interval:
For x = -3:
f(-3) = 3 - 3(-3)^2
= 3 - 3(9)
= 3 - 27
= -24
For x = 2:
f(2) = 3 - 3(2)^2
= 3 - 3(4)
= 3 - 12
= -9
- Evaluate the function at the critical point:
For x = 0:
f(0) = 3 - 3(0)^2
= 3 - 3(0)
= 3 - 0
= 3
Step 3: Compare the values obtained in step 2 to find the absolute maximum and minimum values.
In this case, the absolute maximum value is 3, which occurs when x = 0.
The absolute minimum value is -24, which occurs when x = -3.
Thus, the absolute maximum value of f(x) is 3, and the absolute minimum value is -24 in the interval -3 <= x <= 2.