# local max/min and inflection point

• Jan 26th 2009, 07:18 PM
h4hv4hd4si4n
local max/min and inflection point
The derivative of a function is f'(x)= (x-1)²(x+3).
Find the value of x at each point where f has a
(a) local maximum
(b) local minimum
(c) point of inflection

I got that the maximum is at x=-3, and minimum at x=1. I got x=-5/3 and x=1 as points of inflection.
Can anyone tell me if I'm right, please?
• Jan 26th 2009, 07:36 PM
GaloisTheory1
Quote:

Originally Posted by h4hv4hd4si4n
The derivative of a function is f'(x)= (x-1)²(x+3).
Find the value of x at each point where f has a
(a) local maximum
(b) local minimum
(c) point of inflection

I got that the maximum is at x=-3, and minimum at x=1. I got x=-5/3 and x=1 as points of inflection.
Can anyone tell me if I'm right, please?

solve where derivative is zero and then check the signs on different intervals
• Jan 26th 2009, 08:39 PM
mollymcf2009
The answers you have are for the graph of f'. You use f' to find where f is increasing and decreasing. If you put the graph of f' into your graphing calculator, you will see that it is negative (below the x axis) from - infinity to -3 and positive (above the x axis) from -3 to infinity. What this tells you is that the graph of f is decreasing (could be negative or positive) to the left of -3 and increasing (negative or positive) from -3 to 1 and also to the right of 1.

A minimum or maximum on the graph of f only occurs if there is a change from decreasing to increasing or vice versa. Since you only have one change of sign (which occurs when the derivative =0) this will be your local minimum. The graph of f does not have a local maximum. Now, f' does = 0 at x=1, but notice in the equation of f' that this zero has a power of 2. This tells you that it does equal zero at 1, but it only touches the x-axis, it does not cross it. So, there is no sign change at x=1 and therefore is not a candidate for max or min.