Show that x^5-5x^4+5x^3-1

Question

is a max at x=1,a min at x=3?

Answer 1

The expression is x⁵ - 5x⁴ + 5x³ - 1.

Let the function is f(x) = x⁵ - 5x⁴ + 5x³ - 1.

f  ^' (x) = 5x⁴ - 20x³ + 15x²

f  ^'' (x) = 20x³ - 60x² + 30x

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

5x⁴ - 20x³ + 15x² = 0

5x²(x² - 4x + 3) = 0

x = 0 and x² - 4x + 3 = 0

x² - 4x + 3 = 0

x² - 3x - x + 3 = 0

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

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

The critical numbers are x = 0, x = 1 and x = 3.

If f " (c) > 0 (positive) ------> minimum point,.

If f " (c) < 0 (negative) ------> maximum point.

So, lets plug each critical point in f  ^'' (x) = 20x³ - 60x² + 30x.

At  x = 0

f  ^'' (0) = 20(0)³ - 60(0)² + 30(0)

         = 0 - 0 + 0

            = 0 (Test fails).

At  x = 1

f  ^'' (1) = 20(1)³ - 60(1)² + 30(1)

         = 20 - 60 + 30

         = - 10 < 0

         = local minimum

At  x = 1

f  ^'' (3) = 20(3)³ - 60(3)² + 30(3)

         = 540 - 540 + 30

         = 30 > 0

         = local minimum.

Yes, a maximum at x = 1 and minimum at x = 3.