What is the extreme f(x)=102t+0.5t^2-t^3 ?

The extreme f(x)=102t+0.5t^2-t^3 is Minimum(-17/3 ,-379.98148…),Maximum(6,414)

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extreme f(x)=102t+0.5t^2-t^3

extreme points

f (x) = 102 t + 0.5 t^{2} - t^{3}

Minimum (- \frac{17}{3}, - 379.98148 \dots), Maximum (6, 414)

Solution steps

f^{'} (t) = 102 + t - 3 t^{2}

Find intervals:

Decreasing

: - \infty < t < - \frac{17}{3},

Increasing

: - \frac{17}{3} < t < 6,

Decreasing

: 6 < t < \infty

Plug

t = - \frac{17}{3}

into

102 t + 0.5 t^{2} - t^{3} : - 379.98148 \dots

Minimum (- \frac{17}{3}, - 379.98148 \dots)

Plug

t = 6

into

102 t + 0.5 t^{2} - t^{3} : 414

Maximum (6, 414)

Minimum (- \frac{17}{3}, - 379.98148 \dots), Maximum (6, 414)

f (x) = x^{3} - 8 x^{2} - 12 x + 5

f (x) = x^{3} - 8 x^{2} - 12 x + 9

f (x, y) = (2 x - x^{(2)}) (2 y - y^{(2)})

f (x) = 8 + (9 + 7 x)^{\frac{2}{7}}

f (x) = 6 x + 7 y

What is the extreme f(x)=102t+0.5t^2-t^3 ?
The extreme f(x)=102t+0.5t^2-t^3 is Minimum(-17/3 ,-379.98148…),Maximum(6,414)