Max/Min and Critical Points

Question

 

 

QUESTION TWO

 

Answer 1

(2).(a).

The function is f(x) = 3x⁴ - 4x³ + 2 and interval is [-1, 2].

f^,(x) = 12x³ - 12x².

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

12x³ - 12x² = 0

12x²(x - 1) = 0

The critical numbers are x = 0 and x = 1. So, the number of critical points is 2.

	x - value	f(x) = 3x4 – 4x3 + 2	Conclusion
Left Endpoint	-1	f(-1) = 3(-1)4 – 4(-1)3 + 2 = 3 + 4 + 2 = 9
Critical number	0	f(0) = 3(0)4 – 4(0)3 + 2 = 0 - 0 + 2 = 2
Critical number	1	f(1) = 3(1)4 – 4(1)3 + 2 = 3 - 4 + 2 = 1	Minimum
Right Endpoint	2	f(2) = 3(2)4 – 4(2)3 + 2 = 48 - 32 + 2 = 18	Maximum

The critical numbers are x = 0 and x = 1. The minimum value is f(1) = 1 and the maximum value is f(2) = 18.

(2).(b).

The function is f(x) = (x² - 1)³ and interval is [-2, 2].

f ' (x) = 6x(x² - 1)².

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

6x(x² - 1)² = 0

The critical numbers are x = 0 and x = ± 1. So, the number of critical points is 3.

	x - value	f(x) = (x2 – 1)3	Conclusion
Left Endpoint	- 2	f(- 2) = [(- 2)2 – 1]3 = 33 = 27	Maximum
Critical number	-1	f(- 1) = [(- 1)2 – 1]3 = 03 = 0
Critical number	0	f(0) = [(0)2 – 1]3 = (-1)3 = - 1	Minimum
Critical number	1	f(1) = [(1)2 – 1]3 = 03 = 0
Right Endpoint	2	f(2) = [(2)2 – 1]3 = 33 = 27	Maximum

The critical numbers are x = - 1, x = 0 and x = 1. The minimum value are is f(0) = -1 and the maximum values are f(-2) = 27 and f(2) = 27.

 

 

Answer 2

(1).

The function is Y = kN/(36 + N²), where Y = yield and N = nitrogen level.

Derivative respect to 'N' to each side.

dY/dN = [(36 + N²)k - 2kN²] / (36 + N²)²

dY/dN = k(36 - N²) / (36 + N²)²

Derivative respect to 'N' to each side.

d²Y/dN² = [-2Nk(36 + N²)² - 4Nk(36 - N²)(36 + N²)] / (36 + N²)⁴

              = [-2Nk(36 + N²)²{36 + N² + 72 - 2N²}] / (36 + N²)⁴

              = -2Nk(108 - N²) / (36 + N²)²

To find the critical or key numbers, to make the first derivative equal to zero or dY/dN = 0.

k(36 - N²) / (36 + N²)² = 0

(6² - N²) = 0

The critical numbers are N = 6 and N = -6. So, the number of critical points is 2.

At N = 6

d²Y/dN² = -2(6)k(108 - 6²) / (36 + 6²)²

             = - 864k/5184 < 0 (negative)                [k = positive]

             = Local maximum.

The best yield at N = 6.