Analyze the polynomial function f(x)=x^2(x-6) using parts (a) through (e). Question content area bottom Determine the maximum number of turning points on the graph of the function.
(a) To find the turning points of the function, we need to take the derivative of the function f(x) and set it equal to zero:
f'(x) = 2x(x-6) + x^2(1) = 2x^2 - 12x +x^2 = 3x^2 - 12x
Setting f'(x) = 0, we get:
3x^2 - 12x = 0
3x(x-4) = 0
x = 0 or x = 4
Therefore, the function has two potential turning points at x = 0 and x = 4.
(b) To determine the nature of the turning points, we need to find the second derivative of the function:
f''(x) = d/dx [3x^2 - 12x] = 6x - 12
At x = 0, f''(0) = -12 < 0, so the turning point at x = 0 is a local maximum.
At x = 4, f''(4) = 6(4) - 12 = 24 - 12 = 12, which is greater than 0. Therefore, the turning point at x = 4 is a local minimum.
(c) Therefore, the function f(x) = x^2(x-6) has one local maximum at x = 0 and one local minimum at x = 4.
(d) The maximum number of turning points on the graph of the function is 1, since there is only one local maximum and one local minimum.
(e) The graph of the function f(x) = x^2(x-6) will have one turning point, which is a local maximum at x = 0 and a local minimum at x = 4.