What is the extreme y= 1/9 [(t^3)/3+(3t^2)/2-28t] ?

The extreme y= 1/9 [(t^3)/3+(3t^2)/2-28t] is Maximum(-7, 931/54),Minimum(4,-200/27)

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extreme y= 1/9 [(t^3)/3+(3t^2)/2-28t]

extreme points

y = \frac{1}{9} [\frac{t ^{3}}{3} + \frac{3 t ^{2}}{2} - 28 t]

Maximum (- 7, \frac{931}{54}), Minimum (4, - \frac{200}{27})

Solution steps

f^{'} (t) = \frac{1}{9} (t^{2} + 3 t - 28)

Find intervals:

Increasing

: - \infty < t < - 7,

Decreasing

: - 7 < t < 4,

Increasing

: 4 < t < \infty

Plug

t = - 7

into

\frac{1}{9} [\frac{t ^{3}}{3} + \frac{3 t ^{2}}{2} - 28 t] : \frac{931}{54}

Maximum (- 7, \frac{931}{54})

Plug

t = 4

into

\frac{1}{9} [\frac{t ^{3}}{3} + \frac{3 t ^{2}}{2} - 28 t] : - \frac{200}{27}

Minimum (4, - \frac{200}{27})

Maximum (- 7, \frac{931}{54}), Minimum (4, - \frac{200}{27})

2 x^{3} + 3 x^{2} - 36 x + 3, - 3 \leq x \leq 6

x^{2} + 6 x + 3

f (x) = 2 + 4 x^{2} - x^{4}

\frac{2 x + 8}{x ^{2} + 8}

g (t) = - 5 t^{2} - 3 t + 2

What is the extreme y= 1/9 [(t^3)/3+(3t^2)/2-28t] ?
The extreme y= 1/9 [(t^3)/3+(3t^2)/2-28t] is Maximum(-7, 931/54),Minimum(4,-200/27)