Question 1:
For the function f(x) = (x + 3)(x  2)^{ 3} find
a) the interval(s) of increase and decrease of f,
b) the value(s) of x for which f has a local maximum and minimum,
c) and the interval of concavity and inflection point(s).
Solution to Question 1:

a) We first find the first derivative f '
f '(x) = (x  2)^{ 3} + 3(x + 3)(x  2)^{ 2}
= (x  2)^{ 2}(4x + 7)

f'(x) has zeros at
are x = 2 and x =  7/4

The table of sign of f '(x) is shown below

Using the table, f is decreasing on the interval
(infinity ,  7/3)

and increasing on the interval
( 7/3 , +infinity)

b) f has a local minimum at x =  7/3. Note that although f '(2) = 0, there is no local extremum at x = 2. This is because f ' (x) does not change sign at x = 2.

c) We now calculate f"(x)
f"(x) = 2(x  2)(4x + 7) + 4(x  2)^{ 2}
= 6(x  2)(2x + 1)

The sign table of f"(x)is shown below

Using the above table, we can say that the graph of f is concave up on the intervals (infinity , 1/2) and (2 , +infinity) and concave down on the interval (1/2 , 2).
Question 2:
Given f(x) = e^{ (x 2  x)}, find
a) the interval(s) of increase and decrease of f,
b) value(s) of x for which f has a local extremum,
c) the interval(s) of concavity and any inflection point(s).
Solution to Question 2:

Find f '(x)
a) f '(x) = (2x  1) e^{ (x 2  x)}

f '(x) has one zero at x = 1/2. The sign table of f '(x) is shown below

f is decreasing on (infinity , + 1/2) and increasing on (1/2 , +infinity).

b) Since x = 1/2 is a critical value and f ' changes sign at x = 1/2, there is a local minimum at x = 1/2.

c) We first calculate the second derivative f" of f.
f"(x) = 2 e^{ (x 2  x)} + (2x1)(2x1)e^{ (x 2  x)}
= [ 2 + (2x+1)^{2} ] e^{ (x 2  x)}

The second derivative f" is always positive. Hence the graph of f is concave up on (infinity , +infinity) and since f" does not change sign, the graph of f has no point of inflection.
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