What is the extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} ?

The extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} is Maximum(-4.94444…,-42.06240…)

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extreme f(x)=(8-9x)/(\sqrt[3]{x+3)}

extreme

Maximum (- 4.94444 \dots, - 42.06240 \dots)

Solution steps

Find the critical points:

x \approx - 4.94444 \dots

Domain of

\frac{8 - 9 x}{3 x + 3} : x < - 3 or x > - 3

Find intervals:

Increasing

: - \infty < x < - 4.94444 \dots,

Decreasing

: - 4.94444 \dots < x < - 3,

Decreasing

: - 3 < x < \infty

Plug the extreme point

x = - 4.94444 \dots

into

\frac{8 - 9 x}{3 x + 3} \Rightarrow y = - 42.06240 \dots

Maximum (- 4.94444 \dots, - 42.06240 \dots)

What is the extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} ?
The extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} is Maximum(-4.94444…,-42.06240…)