Maximum and minimum questions

Question

 

QUESTION TWO

Answer 1

(1).

The function is f(x) = 7 + 3x² - 5x³.

f^,(x) = 6x - 15x².

f^,,(x) = 6 - 30x.

Find Extrema :

To find out extrema, use theorem.

If f " (x) > 0 (positive) ------>  Local minimum.

If f " (x) < 0 (negative) ------> Local maximum.

To find the critical or key numbers, to make the first derivative equal to zero or f ' (x) = 0.

6x - 15x² = 0

3x(2 - 5x) = 0

The key numbers are x = 0 and x = 2/5.

So, lets plug each critical point in f " (x) = 6 - 30x.

If x = 0 then f " (0) = 6 - 30(0) = 6 > 0 (positive), local minimum.

If x = 2/5 then f " (2/5) = 6 - 30(2/5) = - 6 < 0 (negative), local maximum.

 

To find the f(x) to each x for local max and local min plugging those values in the original function.

(a). 

At x = 0

f(0) = 7 + 3(0)² - 5(0)³

     = 7 + 0 - 0

      = 7.

The local minimum point is (0, 7) and minimum value of f(x) = 7 at x = 0.

(b).

At x = 2/5

f(2/5) = 7 + 3(2/5)² - 5(2/5)³

         = 7 + 12/25 - 8/25

          = (175 + 12 - 8)/25

          = 179/25

           = 7.16.

The local maximum point is (2/5, 7.16) and minimum value of f(x) = 7.16 at x = 2/5 = 0.4.

 

Answer 2

(2).

The second drivative of f(x) is f^,,(x) = (1-3x)/(1-x)⁴ and its critical numbers are x = 0, x = 2 and x = 1/3.

Find Extrema :

To find out extrema, use theorem.

If f " (x) > 0 (positive) ------> Local minimum.

If f " (x) < 0 (negative) ------> Local maximum.

If f " (x) = 0                 ------> Point of inflection.

 

(a).

At x = 0

f^,,(0) = [1-3(0)]/[1-(0)]⁴

        = 1 > 0 (positive)

        = Local minimum.

The first option is correct .

(b).

At x = 2

f^,,(2) = [1-3(2)]/[1-(2)]⁴

        = - 5 < 0 (negative)

        = Local maximum.

The second option is correct .

 

(c).

At x = 1/3

f^,,(1/3) = [1-3(1/3)]/[1-(1/3)]⁴

           = 0

           = Point of inflection.

The third option is correct .