# `g(x) = x^2 - 2x - 8` Determine the open intervals on whcih the graph is concave upward or downward.

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Given: `g(x)=x^2-2x-8`

Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s).

`g'(x)=2x-2`

`g''(x)=2`

The critical value for the second derivative does not exist.

A critical value will exist for the first derivative.

`g'(x)=2x-2=0`

`2x=2`

`x=1`

If g'(x)>0, the function is increasing in the interval.

If g'(x)<0, the function is decreasing in the interval.

Choose a value for x that is less than 1.

g'(0)=-2 Since g'(0)<0 the function is decreasing in the (-oo,1).

Choose a value for x that is greater than 1.

g'(2)=2 Since g'(2)>0 the function is increasing in the interval (1,`oo).`

Because the function changed direction from decreasing to increasing there exists relative minimum at x=1 and the function is concave up in the interval

(-`oo,oo).`

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