What is the extreme f(x)= 2/3 x^3-7/2 x^2+3x+16 ?

The extreme f(x)= 2/3 x^3-7/2 x^2+3x+16 is Maximum(1/2 , 401/24),Minimum(3, 23/2)

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extreme f(x)= 2/3 x^3-7/2 x^2+3x+16

extreme

Maximum (\frac{1}{2}, \frac{401}{24}), Minimum (3, \frac{23}{2})

Solution steps

f^{'} (x) = 2 x^{2} - 7 x + 3

Find intervals:

Increasing

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

Decreasing

: \frac{1}{2} < x < 3,

Increasing

: 3 < x < \infty

Plug

x = \frac{1}{2}

into

\frac{2}{3} x^{3} - \frac{7}{2} x^{2} + 3 x + 16 : \frac{401}{24}

Maximum (\frac{1}{2}, \frac{401}{24})

Plug

x = 3

into

\frac{2}{3} x^{3} - \frac{7}{2} x^{2} + 3 x + 16 : \frac{23}{2}

Minimum (3, \frac{23}{2})

Maximum (\frac{1}{2}, \frac{401}{24}), Minimum (3, \frac{23}{2})

What is the extreme f(x)= 2/3 x^3-7/2 x^2+3x+16 ?
The extreme f(x)= 2/3 x^3-7/2 x^2+3x+16 is Maximum(1/2 , 401/24),Minimum(3, 23/2)