Determine the max & min of g(x) = x(x^2 - x^3)^3
Get the derivative of the function and set it to zero, then solve for x.
To determine the maximum and minimum of a function, we need to find the critical points. The critical points occur where the derivative of the function is either equal to zero or does not exist.
Let's start by finding the derivative of g(x) = x(x^2 - x^3)^3 with respect to x. We can use the product rule and the chain rule to differentiate it:
g'(x) = (x^2 - x^3)^3 + x * 3(x^2 - x^3)^2 * (2x - 3x^2)
= (x^2 - x^3)^2 * [(x^2 - x^3) + 3x(2x - 3x^2)]
= (x^2 - x^3)^2 * (x^2 - x^3 + 6x^2 - 9x^3)
= (x^2 - x^3)^2 * (7x^2 - 10x^3)
Next, we set the derivative equal to zero and solve for x:
g'(x) = 0
(x^2 - x^3)^2 * (7x^2 - 10x^3) = 0
This equation has two factors, so we can set each factor equal to zero:
x^2 - x^3 = 0 (Factor 1)
7x^2 - 10x^3 = 0 (Factor 2)
To find the solutions, we can solve each equation separately:
Factor 1: x^2 - x^3 = 0
x^2(1 - x) = 0
This equation has two solutions:
x^2 = 0 => x = 0
1 - x = 0 => x = 1
Factor 2: 7x^2 - 10x^3 = 0
x^2(7 - 10x) = 0
This equation has two solutions:
x^2 = 0 => x = 0
7 - 10x = 0 => x = 7/10
Now that we have the critical points, we can evaluate the original function g(x) at these points and determine the maximum and minimum values.
Let's substitute the critical points into the original function:
g(0) = 0(0^2 - 0^3)^3 = 0
g(1) = 1(1^2 - 1^3)^3 = 0
g(7/10) = (7/10)((7/10)^2 - (7/10)^3)^3 = (7/10)(49/100 - 343/1000)^3 = (7/10)(161/1000)^3
Therefore, g(0) = 0 and g(1) = 0. To determine the value of g(7/10), we need to simplify the expression further.
g(7/10) = (7/10)(161/1000)^3
= (7/10)(161/1000)(161/1000)(161/1000)
= 16907/1,000,000
Hence, the maximum value of g(x) is 0 and the minimum value is 16907/1,000,000.