Find the absolute max and min values

Question

Find the absolute maximum and absolute minimum values of the function
f(x)=x^3−12x^2−27x+10
over each of the indicated intervals.

 

a) Interval [-2,0]

1. Absolute max=

2. Absolute min=

 

b) Interval [1,10]

3. Absolute max=

4. Absolute min=

 

c) Interval [-2,10]

5. Absolute max=

6. Absolute min=

 

Answer 1

Set the derivative of the function equal to zero.

 

f'(x) = 0

 

3x² - 24x - 27 = 0

 

(3x + 3)(x - 9) = 0

 

3(x + 1)(x - 9) = 0

 

x = -1      and    x = 9

 

Both critical values are in the interval given.

 

By now, you should recognize the shape of a positive cubic graph.  It has one maximum and one minimum.  The lesser x value has the maximum, and the greater x value has the minimum.  Therefore,

 

f(-1) is a maximum

f(9) is a minimum

Answer 2

 a) Interval [-2,0] find absolute maximum and minimum

 

f'(x) = 3 x² - 24 x -27 is the slope of curve. find the critical point over the interval [-2,0], set  f'(x) = 0

 

f'(x) = 3 x² - 24 x - 27 = 0 , solve quadratic equation and find the roots:

 

divide by 3 both sides , x² - 8x - 9 = 0 , factoring (x+1)(x-9)=0 , x = -1 , and  x = 9 , x = 9 is outside the interval and 

 

will not be considered. The test points are the end of the interval points -2, 0, and -1 which is within the interval 

 

The values of the first and second derivatives at x=-1 are:

 

f'(-1) = 3(-1)² -24(-1)-27 = 0 , f'''(-1) = 6(-1)-24 = -30 < 0  is concave down is a relative maximum for now. It will be compared with the end points to determine if it is a relative or absolute maximum

 

find the value of the function at the end interval points:

 

f(0) = (0)³ - 12 (0)² -27(0) + 10 = 10 

 

f(-2) = ( -2)³ -12 ( -2)² -27(-2) + 10 = 8

 

f(-1) = (-1)³ -12(-1)² -27(-1) + 10 =  24 > f ( 0) = 10 > f(-2) = 8 

 

Therefore, the point x = -2 is an absolute minimum over the interval [-2,0]

                                  

                                  x = -1 is an absolute maximum over the interval [-2,0]

 

Now you can solve sections b) and c) in the same manner....

 

 

 

 

 

Answer 3

First let's get the local maximums over the real number line.

 

The the critical points of y = ax³ + bx² + cx + d are at

 

x = [-b ±√(b² - 3ac)] / (3a).

 

Note that it looks almost identical to the quadratic formula. There is a reason for this.

 

For your cubic, x³ - 12x² - 27x + 10, this is:

 

x = [12 ± √(12² - 3·1·(-27))] / (3·1) = (12 ± 15) / 3 ∈ {9,-1}.

 

The y coordinates are:

 

y(9) = 9³ - 12·9² - 27·9 + 10 = -476  (Local Minimum)

 

y(-1) = (-1)³ - 12·(-1)² - 27·(-1) + 10 = 24 (Local Maximum)

 

 

Thus:

 

a) Interval [-2,0]

 

y(-2) = (-2)³ - 12·(-2)² - 27·(-2) + 10 = 8 

y(-1) = 24

y(0) = 10

 

1. Absolute max = 24
2. Absolute min = 8
 
b) Interval [1,10]

 

y(1) = 1³ - 12·1² - 27·1 + 10 = -28 
y(9) = -476
y(10) = 10³ - 12·10² - 27·10 + 10 = -460

 

3. Absolute max = -28
4. Absolute min = -476
 
c) Interval [-2,10]

 

y(-2) = 8
y(-1) = 24
y(9) = -476

y(10) = -460

 

5. Absolute max = 24
6. Absolute min = -476

Answer 4

Without regard to any restricted interval, the first derivative is set to zero to find the CRITICAL x values of local extrema.

 

Within every closed interval, evaluate the function at both endpoints and at any criticial point falling within the interval.

Choose the lowest of these evaluations as the absolute minimum; choose the highest of these evaluations as the absolute minimum.

 

Repeat for each interval.

 

You CAN AND SHOULD DO THIS.