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

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

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

extreme points

f (x) = \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})

f (x, y) = 3 x + 4 x^{2} y^{3} - 5 y^{2}

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

5 x - 3

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

8 x^{6} - 13 x^{5}

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