What is the extreme f(x)=(x^{5/3})/(2+x),-1<= x<= 8 ?

The extreme f(x)=(x^{5/3})/(2+x),-1<= x<= 8 is Minimum(-1,-1),Saddle(0,0),Maximum(8, 16/5)

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extreme f(x)=(x^{5/3})/(2+x),-1<= x<= 8

extreme

Minimum (- 1, - 1), Saddle (0, 0), Maximum (8, \frac{16}{5})

Solution steps

Find the critical points:

x = - 1, x = 0, x = 8

Domain of

\frac{x ^{\frac{5}{3}}}{2 + x}, - 1 \leq x \leq 8 : - 1 \leq x \leq 8

Find intervals:

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < 8

Plug the extreme point

x = - 1

into

\frac{x ^{\frac{5}{3}}}{2 + x} \Rightarrow y = - 1

Minimum (- 1, - 1)

Plug the extreme point

x = 0

into

\frac{x ^{\frac{5}{3}}}{2 + x} \Rightarrow y = 0

Saddle (0, 0)

Plug the extreme point

x = 8

into

\frac{x ^{\frac{5}{3}}}{2 + x} \Rightarrow y = \frac{16}{5}

Maximum (8, \frac{16}{5})

Minimum (- 1, - 1), Saddle (0, 0), Maximum (8, \frac{16}{5})

What is the extreme f(x)=(x^{5/3})/(2+x),-1<= x<= 8 ?
The extreme f(x)=(x^{5/3})/(2+x),-1<= x<= 8 is Minimum(-1,-1),Saddle(0,0),Maximum(8, 16/5)