# Find the absolute max and min with the local max and min values of f 1. f(x) = 3x2 0 < x < 2 2. f(x) = 7ex 3.

*print*Print*list*Cite

### 1 Answer

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.