Calculus Question

Question

Use the second derivative test to show that (3,27) is a maximum point on the graph of f(x) = x³(4-x).

 

Help!?

Answer 1

f(x) = x³(4-x) = 4x³ - x⁴

f'(x) = 12x² - 4x³

 

Set f'(x) to zero to find the extreme points:

 

     0 = 12x² - 4x³

     4x³ = 12x²

     x = 3

 

So there is an extreme point at x = 3.  The y value is f(3) = 3³(4-3) = 27.  So the extreme point is (3, 27).  To find out if it is a maximum or minimum, take the second derivative and evaluate it at x = 3.  If f''(3) > 0, it's a minimum. If f''(3) < 0, it's a maximum

 

     f''(x) = 24x - 12x²

     f''(3) = 24·3 - 12·3²

     f''(3) = 72 - 108 = -36

 

Since f''(3) < 0, (3,27) is a maximum.

Answer 2

f(x) = 4x³ - x⁴

 

f'(x) = 12x² - 4x³

 

f"(x) = 24x - 12x²

 

f"(x) = 12x(2 - x)

 

12x(2 - x) = 0

 

x = 0           and          x = 2

 

These are inflection points.

 

Evaluate f"(3).  If the second derivative is negative, then f(3) is a maximum.  This is because concave down symbolizes a maximum.

 

f"(3) = 12(3)(2 - 3) = 36(-1) = -36

 

It is negative, therefore, f(3) is a maximum.