What is the extreme R(M)=M^2(1/2-M/3) ?

The extreme R(M)=M^2(1/2-M/3) is Minimum(0,0),Maximum(1, 1/6)

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extreme R(M)=M^2(1/2-M/3)

extreme

Minimum (0, 0), Maximum (1, \frac{1}{6})

Solution steps

f^{'} (M) = M (1 - M)

Find intervals:

Decreasing

: - \infty < M < 0,

Increasing

: 0 < M < 1,

Decreasing

: 1 < M < \infty

Plug

M = 0

into

M^{2} (\frac{1}{2} - \frac{M}{3}) : 0

Minimum (0, 0)

Plug

M = 1

into

M^{2} (\frac{1}{2} - \frac{M}{3}) : \frac{1}{6}

Maximum (1, \frac{1}{6})

Minimum (0, 0), Maximum (1, \frac{1}{6})

What is the extreme R(M)=M^2(1/2-M/3) ?
The extreme R(M)=M^2(1/2-M/3) is Minimum(0,0),Maximum(1, 1/6)