Function defined only on left side of critical point $$x=c$$:
$$\bullet$$If $$f_{-}\left(c\right)0$$ to the left of $$x=c$$
then $$x=c$$ is a local maximum.
$$\bullet$$If $$f_{-}\left(c\right)>f\left(c\right)$$ or $$f_{-}\left(c\right)=f\left(c\right)$$ and $$f\:{^{\prime}}\left(x\right)<0$$ to the left of $$x=c$$
then $$x=c$$ is a local minimum.

Function defined only on right side of critical point $$x=c$$:
$$\bullet$$If $$f_{+}\left(c\right)then $$x=c$$ is a local maximum.
$$\bullet$$If $$f_{+}\left(c\right)>f\left(c\right)$$ or $$f_{+}\left(c\right)=f\left(c\right)$$ and $$f\:{^{\prime}}\left(x\right)>0$$ to the right of $$x=c$$
then $$x=c$$ is a local minimum.

Function is discontinous and defined on both side of critical point $$x=c$$:
$$\bullet$$If $$f_{-}\left(c\right)0$$ to the left of $$x=c$$
and $$f_{+}\left(c\right)then $$x=c$$ is a local maximum.
$$\bullet$$If $$f_{-}\left(c\right)>f\left(c\right)$$ or $$f_{-}\left(c\right)=f\left(c\right)$$ and $$f\:{^{\prime}}\left(x\right)<0$$ to the left of $$x=c$$
and $$f_{+}\left(c\right)>f\left(c\right)$$ or $$f_{+}\left(c\right)=f\left(c\right)$$ and $$f\:{^{\prime}}\left(x\right)>0$$ to the right of $$x=c$$
then $$x=c$$ is a local minimum.
$$\bullet$$If $$f_{-}\left(c\right)0$$ to the left of $$x=c$$
and $$f_{+}\left(c\right)>f\left(c\right)$$ or $$f_{+}\left(c\right)=f\left(c\right)$$ and $$f\:{^{\prime}}\left(x\right)>0$$ to the right of $$x=c$$
then $$x=c$$ is neither a local maximum nor a local minimum.
$$\bullet$$If $$f_{-}\left(c\right)>f\left(c\right)$$ or $$f_{-}\left(c\right)=f\left(c\right)$$ and $$f\:{^{\prime}}\left(x\right)<0$$ to the left of $$x=c$$
and $$f_{+}\left(c\right)then $$x=c$$ is neither a local maximum nor a local minimum.

extreme f(x)=5-8x,x>= 1

Solution

Solution

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What is the extreme f(x)=5-8x,x>= 1 ?