the number of inflection points for the function: x^3-4x^2 is what? please help, i keep getting one inflection

Answer 1

f(x) = x^3 - 4x²
f'(x) = 3x² - 4x
f''(x) = 6x - 4

at x = 2/3, f''(x) changes sign. This makes it an inflection point. Left of there, the curve is concave down, right of there it's concave up. But at x = 0 and x = 4/3, f'(x) = 0. These are extrema, but perhaps your instructor counts these as inflection points, too. They are certainly "critical points" where interesting things are going on with the function. Since f(x) is concave down on (-∞, 2/3), x=0 is a local maximum, and since f(x) is concave up on (2/3, ∞), x = 4/3 is a local minimum. This information alone should be enough to sketch a graph.

Answer 2

You keep getting only one inflection point because there is only one:
y = x^3 - 4x^2
y' = 3x^2 - 8x
y'' = 6x - 8 = 0
x = 4/3

check:
y'(1) = 3 - 8 = -5
y'(4/3) = 3*16/9 - 8*4/3 = 16/3 - 32/3 = -16/3
y'(5/3) = 3*25/9 - 8*5/3 = -5
so the slope increases until x = 4/3, then begins to decrease

Answer 3

f' is 3x^2 - 8x

f'' is 6x -8

thus it is concave down at x < 8/6
and concave up at x > 8/6, thus one inflection point because f'' changes form - to +