N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88 -------(i)

That is a cubic function. If you graph that, you'd see that it is like a strectched slightly vertical S. Meaning, the graph goes up, goes down, then goes up again.

One of the characteristics of the cubic curve is it goes upward very fast, slows down to stop at the relative maximum point, goes downward, picking speed downward until it wants to go upward again. This is at the inflection point. This is where the graph attains its maximum downward speed. Then passing that inflection point, the graph will change curvature, preparing to go upward again. So it slows down on its downward plunge until it reaches the relative minimum point, after which it shoots up for the sky, picking up speeds as time goes by, because it won't go downward anymore.

So,

a)The graph goes downward fastest at the inflection point. We can get that through the second derivative of N(t). At the inflection point, N''(t) = 0.

b) maximum speed upward, in the range of the function, is either at the initial point, at t=0, or at terminal point, at t=39. We can get the rates through the first derivatve of N(t)

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N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88 -------(i)

N'(t) = 0.084702t^2 -2.1844t +13.029

N''(t) = 0.169404t -2.1844

Set that to zero,

t = 2.1844 /0.169404 = 12.89 or 13 years after 1950.

Therefore, the prison population was decreasing most rapidly in 1963. ----answer.

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N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88 -------(i)

N'(t) = 0.084702t^2 -2.1844t +13.029

N'(0) = +13.029

N'(39) = 0.084702(39^2) -2.1844(39) +13.029 = +56.67

So, the slope of the curve is greater at t=0 than at t=39.

Therefore, the prison population was increasing most rapidly in 1950. ----answer.