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Mathematics 18 Online
OpenStudy (anonymous):

How can I find critical points using a derivative?

OpenStudy (anonymous):

first derivative must be equal to 0 at a critical point

OpenStudy (amistre64):

.. or undefined

OpenStudy (amistre64):

since the derivative tells us the slope of the tangent line to the curve; and we know that a slope of 0 is a horizontal line; this is a point of interest.

OpenStudy (anonymous):

And how do I determine which type of critical point it is?

OpenStudy (amistre64):

where the slope is undefined; we have points that need to be checked since they can be cusps or corners or other things that simply have no derivative to speak of

OpenStudy (amistre64):

by going a little to the left and right of it to see if the slope changes direction of continues on its happy little way

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

if it changes direction, your at a min or max; of it continues on; your most likely at an inflection point

OpenStudy (anonymous):

with my specific equation f(x) = X^2-6X+1, there is only one critical point, and i find it by taking the first derivative and then testing points to either side of the point where the derivative equals 0, correct?

OpenStudy (amistre64):

yes. that is one way to check it. knowing that it is a parabola helps

OpenStudy (anonymous):

yep. thank you very much

OpenStudy (amistre64):

your welcome :)

OpenStudy (anonymous):

when i have a function with more degrees, i simply continue to derive the derivatives, correct?

OpenStudy (amistre64):

you can, but you should always test out the critical points to see what they act like. A value of 0 on the derivatives is a good indication, but can be a false reading. Also, points of interest can be when it goes undefined

OpenStudy (anonymous):

okay. would it help to graph the function to see what to expect?

OpenStudy (amistre64):

oh, and end points ;)

OpenStudy (amistre64):

yes, graphing is always a good visual; but not always practical

OpenStudy (anonymous):

thank you so so much :D

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