What is the extreme f(x)=(2x^3)/3+(7x^2)/2-4x ?

The extreme f(x)=(2x^3)/3+(7x^2)/2-4x is Maximum(-4, 88/3),Minimum(1/2 ,-25/24)

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extreme f(x)=(2x^3)/3+(7x^2)/2-4x

extreme points

f (x) = \frac{2 x ^{3}}{3} + \frac{7 x ^{2}}{2} - 4 x

Maximum (- 4, \frac{88}{3}), Minimum (\frac{1}{2}, - \frac{25}{24})

Solution steps

f^{'} (x) = 2 x^{2} + 7 x - 4

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - 4

into

\frac{2 x ^{3}}{3} + \frac{7 x ^{2}}{2} - 4 x : \frac{88}{3}

Maximum (- 4, \frac{88}{3})

Plug

x = \frac{1}{2}

into

\frac{2 x ^{3}}{3} + \frac{7 x ^{2}}{2} - 4 x : - \frac{25}{24}

Minimum (\frac{1}{2}, - \frac{25}{24})

Maximum (- 4, \frac{88}{3}), Minimum (\frac{1}{2}, - \frac{25}{24})

f (x) = \frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x + \frac{4}{3}

- 3 sin (2 x) - 4

f (x) = tan (x), - \frac{π}{6} \leq x \leq \frac{π}{4}

f (x) = x^{4} - 4 x^{3} - 8

x, f = x^{2} - 4 x y + y^{2}

What is the extreme f(x)=(2x^3)/3+(7x^2)/2-4x ?
The extreme f(x)=(2x^3)/3+(7x^2)/2-4x is Maximum(-4, 88/3),Minimum(1/2 ,-25/24)