easy calculus problem (extremas)?
Determine the absolute extremas of the function and the x-value in the closed interval where it occurs
FUNCTION: f(x)= 2(3-x)
INTERVAL: [-1,2]
alright so first i got the derivative of the function.... -2? then i plugged that in and got a 10... spoiler alert that's not the answer. the answer is
Minimum: (2,2)
Maximum: (-1,8)
Comments
Think about it.
f(x) = 6-2x is the graph of a straight line. That's why the derivative is just a horizontal line at -2. The derivative (slope) is always -2.
So, naturally, the extremas would be at the endpoints of the line at x=-1 and x=2. They plugged these numbers in and got the answer.
In order to find absolute extrema the x-values you take into consideration are the end points and where the derivative equals zero
Since the derivative is f'(x)=-2 and -2 does not and cannot equal zero the only values you have to use are the end points -1 and 2
if you plug 2 into f(x) you get 2
if you plug -1 into f(x) you get 8
8>2 so therefore 2 is your absolute min x-value and -1 is your absolute max x-value
hence (2, 2) and (-1, 8)