Find the absolute max and min with the local max and min values of f 1. f(x) = 3x2 0

1) To find the maximum or miminimum for 3x^2 in 0 <
x <2.

Solution:

f(x) =
3x^2.

f'(x) =2x

f'(x) is
> 0 for all x  for which 0 < x< 2.

So
f'(x) is a, continuous increasing function.

Therefore , the
local minimum amd local maximum (also global minimum and global maximum ) are  at 0 and
2. But these points are not  the interior of the given open domain 0 < x <
2. So there is no local or global minimum  for f(x) = 3x^2 in the open  domain 0
< x< 2.

2)

f(x)
= 7e^x

f'(x) = 7e^x >
0.

It is a continuously increasing function and has no
point where it is maximum. For any closed interval  (a,b) or a< = x < =
b it has the local minimum  and global minimum at x= a , and local maximum and global
maximum at b. But for any open interval a < x < b , there is no local or
global minimum or
maximum.

3)

f(x) = 25-x^2 if
-5 <= x < 0

f(x) = 3x -3 if  0 <= 3
< =5.

The critical points in -5<=x < 0
are given by the boundary -5 , the solution of (25-x^2)' = 0 , or  -2x = 0 , x = 0. But
x = 0 is not an interior point.

Since -2x is positive in -5
<= x < 0, f(x) = (25-x^2) is  increasing function  for -5<= x
< 0 which is left closed  but right open. As 0, being exterior point, there is
no maximum. x = -5 is a local minimum, a minimum value of 25-x^2 = 25-(-5)^2 =
0.

At x = 0 , there is a jump. As f(0-) = 25 and f(0+) =
-3.

f(x) = 3x-3 in 0 <= x <
=5.

f'(x) = 3. So this is an increasing function in  the
closed interval (0 , 5)

So in (0 , 5), at  x = 0,  3x-3 = 
3*0-3 = -3 is the local minimum and at x= 5,  3*5-3 = 12 is the local
maximum.

So the global minimum  = minimum  of (0  and  -3)
=  -3 at x = 0 of the domain 0<= x <
=5.

global maximum  of ( local max in -5 < x
< 0  local maximum in 0 <= x <
5).

 But f(0-) = 25  for any point in (sqrt10)-25 <
= x < 0, f(x) > f(5) = 15.

So global maximum
does not exist.