What is the extreme f(x)=4-16/3 x^2,0<= x<= 1/2 ?

The extreme f(x)=4-16/3 x^2,0<= x<= 1/2 is Maximum(0,4),Minimum(1/2 , 8/3)

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extreme f(x)=4-16/3 x^2,0<= x<= 1/2

extreme

Maximum (0, 4), Minimum (\frac{1}{2}, \frac{8}{3})

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}

Domain of

4 - \frac{16}{3} x^{2}, 0 \leq x \leq \frac{1}{2} : 0 \leq x \leq \frac{1}{2}

Find intervals:

Decreasing

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

Plug the extreme point

x = 0

into

4 - \frac{16}{3} x^{2} \Rightarrow y = 4

Maximum (0, 4)

Plug the extreme point

x = \frac{1}{2}

into

4 - \frac{16}{3} x^{2} \Rightarrow y = \frac{8}{3}

Minimum (\frac{1}{2}, \frac{8}{3})

Maximum (0, 4), Minimum (\frac{1}{2}, \frac{8}{3})

What is the extreme f(x)=4-16/3 x^2,0<= x<= 1/2 ?
The extreme f(x)=4-16/3 x^2,0<= x<= 1/2 is Maximum(0,4),Minimum(1/2 , 8/3)