is undefined, we would expect f to have a sharp corner or a vertical
tangent at x=c. We know from previous discussion that the
sharp corner is important because it could create a max or a min for f.
Let's use examples to demonstrate these.
Example 1: Let 
(1) First, we rewrite 
and find 
So the critical number for f is at x=2. When x>2, 
so f is increasing in 
When 
so f is decreasing in 
We would expect f to have a minimum at x=2. Since f(2)=1,
the minimum point for f is (2,1).
(2) We note that 
is undefined (and f(2) is defined) so we expect to have a sharp
corner or a vertical tangent at x=2. But vertical tangent is impossible,
since we already see that f has a min. at x=2. So, it has
to be a sharp corner. Pay attention to the following two questions which
are related to the slope of the tangent at x=2.
(3) What is 
(If you know there is a sharp corner and min. at x=2, you could
guess the answer to this already). Let's evaluate the limit, 
(you plug in x=2.0001), you see the denominator is small, and the
fraction is posititive, so you get 
(4) What is 
Try this for yourself, you should get 
(5) Can you finish the graph? Use plot(((x-2)^(2/3)) + 1,x=-2..4, thickness=3); and use Maple Software to verify your answer, but be cautious, the graph you obtain from Maple is not accurate, because most of computer algebra systems, they have trouble to plot odd root functions. They treat them like even root functions.
(6) Use 
to find the intervals where f is convave upward and downward, from
our graph and analysis, we should see that f is concave downward
in 
Let's see if we get consistent information from 
Well, since 
we see that denomator is always positive so the whole fr action is always
negative, which implies that f is always concave down.
Example 2: Let 
.
Investigate 
max/min, concave upward/downward, sharp corner or v ertical tangent and
graph the function.
(1) First, 
and 
Note that 
is always positive, so f is always increasing, no max or min.
(2) We note that 
is undefined but f(-5) is defined (=0), we expect to have a vertical
tangent at x=-5 (will not be a sharp corner because we should not
have max or min.) Can you guess what 
and 
(3)
(Verify).
(4) Since 
when x<-5, 
so f is concave upward in 
When x>-5, 
so f is concave downward in (-5, 
(5) Use plot(4*((x+5)^(3/5)) + 1,x=-2..7, thickness=3); Notice that Maple gives a wrong graph for this problem, do you know why?