Quote:
Originally Posted by magmagod
Use an algebric strategy to verify that the point given for each function is either a maximum or a minimum.
x^33x ; (1, 2)
so i tried finding the instantaneous rate of change before 1 and after 1 to see if its a maximum or a minimum, since if its a local max before 1 would be positive and after would be negative but i couldn't reach the solution in the back of the book. any help would be appreciated.
the formula i used was the difference quotient.

Well I thought I'd have a go at this one anyway!
Here are my thoughts:
f(x) = x^3  3x; verify point (1,2)
f '(x) = 3x^2 3
For max or min f '(x) = 0 so we get:
3x^2 3 = 0 (divide this though by 3 we get next line)
x^2 1 = 0
then factorising we get
(x + 1) (x  1) = 0
so x = 1 or x = 1
Substituting this back into the first eqn f (x) we can get the Y coordinates and determine whether local max or min.
so f (1) = (1)^3 3(1) = 1 + 3 = 2 so coordinates are (1, 2) = Y max
and f (1) = (1)^3 3(1) = 1 3 = 2 so coordinates are (1 , 2) = Y min
Remember these are local max and mins not the highest or lowest points on the curve of f (x) = x^3  3x.
See diagram of curve from x = 3 to x = 3 for confirmation.
Hope this is of use for someone