anonymous
  • anonymous
How do you find critical points? (get at least 2 of the 3 acceptable answers)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
are you asking a method?
anonymous
  • anonymous
yeah, for example I know that you can find critical points by taking the derivative and setting it equal to zero, but what are the other possible ways?
anonymous
  • anonymous
You can plot a graph and then see the critical points - that's another method.

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More answers

anonymous
  • anonymous
There's one more way, but I cant seem to get it. :/
anonymous
  • anonymous
yes..or plot on a sign diagram...(never liked that way) for example, f(x)=(x-1)^2 factor out f(x)=(x=1)*(x-1), set x+1=0 set x-1=0. these will be crit. points
anonymous
  • anonymous
oops..f(x)=(x+1)*(x-1)
anonymous
  • anonymous
So, to find critical points are you saying we just factor out the function and set it equal to zero (instead of taking the derivative?)
anonymous
  • anonymous
my mistake...been a while. no done for first derivative...
anonymous
  • anonymous
are you trying to get the relative max and min, or also the absolutes?
anonymous
  • anonymous
For this question?
anonymous
  • anonymous
no..just trying to get context, since the 2nd der. will give absolutes
anonymous
  • anonymous
I'm sorry, Im confused.
anonymous
  • anonymous
well..me too. you want 3 ways, in general, to find the critical values?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
the critical points are max, min and points of inflexion.
anonymous
  • anonymous
ok
anonymous
  • anonymous
ok. 2nd derivatives are the inflection points
anonymous
  • anonymous
1. Find the derivative and set it to 0. This way, you get max and min. The second derivative should give you points of inflexion. 2. Plot the equation on a graph and find the max, min and inflexion points visually. 3. Use a computer/calculator ;)?
anonymous
  • anonymous
calculator. hehehe
anonymous
  • anonymous
lol, maybe so. Thanks, guys (:
anonymous
  • anonymous
good luck

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