`f(x) = x(x - 4)^3` Find the points of inflection and discuss the concavity of the graph of the function. f(x) is concave up

First derivative,

`f'(x) = 1*(x-4)^3+x*3*1*(x-4)^2`

`f'(x) = (x-4)^2(x-4+3x)`

`f'(x) = 4(x-4)^2(x-1)`

There are only two critical points, x = 4 and x = 1

Second derivative,

`f''(x) = 4(2*1*(x-4)*(x-1)+ (x-4)^2*1)`

`f''(x) = 4(x-4) (2x-2+x-4)`

`f''(x) = 4(x-4)(3x-6)`

`f''(x) = 12(x-4)(x-2)`

For inflection points, f''(x) = 0.

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f(x) is concave up

First derivative,

`f'(x) = 1*(x-4)^3+x*3*1*(x-4)^2`

`f'(x) = (x-4)^2(x-4+3x)`

`f'(x) = 4(x-4)^2(x-1)`

There are only two critical points, x = 4 and x = 1

Second derivative,

`f''(x) = 4(2*1*(x-4)*(x-1)+ (x-4)^2*1)`

`f''(x) = 4(x-4) (2x-2+x-4)`

`f''(x) = 4(x-4)(3x-6)`

`f''(x) = 12(x-4)(x-2)`

For inflection points, f''(x) = 0.

Therefore, the inflection point is at x = 4, not at x = 1.

x<2

f"(x) >0, therefore f'(x) is increasing, which means f(x) is concave up at x<2.

2<x<4

f"(x) < 0, therefore f'(x) is decreasing, which means f(x) is concave down at 2<x<4

x>4

f"(x) > 0, therefore f'(x) is increasing, which means f(x) is concave up at x>4

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