How Do I Find The Slope Of The Tangent Line From A Critical Pointg(X)=2+4X(4-X) I Solved For The 1St Derivative For The Critical Point, 2Now How Do I Find The Slope Of A Tangent Line From This, And Two Points Where The Average Rate Of Change Is 0?
A critical point is a value for x where g ‘ (x) is zero or where g is not differentiable. So you found the critical point 2. The slope of the tangent line at a point x is the same thing as g ‘ (x) at that point, so you already have the answer to this since you the x you found was the point at which g ‘ (x) was ZERO. “Two points where the average rate of change is zero” – does not make sense. An average has to be over a range. So perhaps you meant two points BETWEEN which the average rate of change is zero. These are any two points at the same height. From the symmetry of the function, x=0 and x=4 look like good candidates. The slope of the line joining these two points, is given by increase in y divided by increase in x.
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