What is the extreme f(x)=345x^2-3450x^3,0<= x<= 0.1 ?

The extreme f(x)=345x^2-3450x^3,0<= x<= 0.1 is Minimum(0,0),Maximum(1/15 , 23/45),Minimum(0.1,0)

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extreme f(x)=345x^2-3450x^3,0<= x<= 0.1

extreme

Minimum (0, 0), Maximum (\frac{1}{15}, \frac{23}{45}), Minimum (0.1, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{1}{15}, x = 0.1

Domain of

345 x^{2} - 3450 x^{3}, 0 \leq x \leq 0.1 : 0 \leq x \leq 0.1

Find intervals:

Increasing

: 0 < x < \frac{1}{15},

Decreasing

: \frac{1}{15} < x < 0.1

Plug the extreme point

x = 0

into

345 x^{2} - 3450 x^{3} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{1}{15}

into

345 x^{2} - 3450 x^{3} \Rightarrow y = \frac{23}{45}

Maximum (\frac{1}{15}, \frac{23}{45})

Plug the extreme point

x = 0.1

into

345 x^{2} - 3450 x^{3} \Rightarrow y = 0

Minimum (0.1, 0)

Minimum (0, 0), Maximum (\frac{1}{15}, \frac{23}{45}), Minimum (0.1, 0)

What is the extreme f(x)=345x^2-3450x^3,0<= x<= 0.1 ?
The extreme f(x)=345x^2-3450x^3,0<= x<= 0.1 is Minimum(0,0),Maximum(1/15 , 23/45),Minimum(0.1,0)