The confidence interval is to .
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I let the computer do the graphing.
f(x) = x(x + 3)²(1 - x) / 3
We see obvious zeros at x=0, x=-3 and x=1.
f(x) = x(x + 3)²(1 - x) / 3
f(x) = x(x^2 + 6x + 9)(1 - x)/3
f(x) = x(x^2 + 6x + 9 - x^3 - 6x^2 - 9x)/3
f(x) = x(- x^3 - 5x^2 -3x +9)/3
f(x) = (-x^4 - 5x^3 -3x^2 +9x)/3
f'(x) = (-4x^3 - 15x^2 -6x + 9)/3
We know there's f'(-3)=0 so x+3 must be a factor here,
f'(x) = (x + 3) (-4x^2 - 3x + 3)
We get stationary points when x = -3 or
x = (-1/8)( 3 ± √(9 - 4(-4)(3))) = (-3 ± √57)/8 about -1.3 and 0.6 which are a local minimum and maximum respectively.
f''(x) = (-12x^2 - 30x -6)/3 = -4x^2 - 10x - 2
f''(-3) = -4(9)+ 30 - 2 = -8
Concave Down, Negative. (-3,0) is a local maximum.
f''((-3 - √57)/8) = -4(-1.3)^2 - 10(-1.3) - 2 = 4.24
Concave up, positive, a CUP
(-3 - √57)/8, f((-3 - √57)/8) ) is a minimum.
f''((-3 + √57)/8) = -4(.6)^2 - 10(.6) - 2 = -9.44
CDN, ( ((-3 + √57)/8), f((-3 + √57)/8) ) is a maximum.
I didn't evaluate f((-3 ±√57)/8) which are around -3 and 1 but I leave that calculation to you.
