Draw the sign diagram for f ' and find the relative extrema of f

timer Asked: Jul 15th, 2015
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School: UT Austin

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1) Let's notice that if we choose any x value less than 3 (e.g. x = 2) and we make a tangent line at x = 2 then we will have a line that is going up (its slope is positive) then this means f '(x<3) = Positive (+).

2) Then if we make a tangent line at x = 3 then we will have a horizontal line (its slope is zero) which means           f '(3) = 0. When this happens then it means that it is a relative maximum or minimum. It is a minimum if it concaves up (like a parabola opens up --> this way U) and it is a maximum if it concaves down (like a parabola opens down). So at x = 3 we have a relative maximum.

3) Between  x = 3 and x = 5 (e.g. at x = 4) we can see that if we make a tangent line at x = 4 then the line goes down (its slope is negative) which means f'(3< x < 5) = Negative (-).

4) At x = 5 the tangent line is horizontal, so f'(5) = 0 and since it is concaves up like a U then we have a relative minimum at x = 5.

5) Between x = 5 and x = 7 we can see that the tangent line is going up. So f'(5< x <7) = Positive (+).

6) At x = 7 we can see a change in its concavity, it concaves down first and then it concaves up and its tangent line is horizontal which means f'(7) = 0 (its slope is 0). So when this happens then it means that it is an inflextion point (it is not a maximum and neither a minimum, it's just an inflexion point). So at x = 7 is an inflexion point and    f'(7) = 0.

7) Between x = 7 and x = 9 the tangent line is going up (its slope is positive). So f'(7 < x < 9) = Positive (+).

8) At x = 9 the tangent line is horizontal. So f'(9) = 0 and it is a relative maximun at x = 9.

9) f'( 9< x < 12) = Negative (-), the tanagent line is going down.

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