jmartinez638
  • jmartinez638
find the local minimum and maximum points and points of inflection for the function f(x) = 1/3X^3 - x^2 - 35x +17
Mathematics
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SOLVED
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chestercat
  • chestercat
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jmartinez638
  • jmartinez638
\[f(x) =\frac{ 1 }{ 3 } x^3 - x^2 - 35x +17\]
SolomonZelman
  • SolomonZelman
What is the first derivative?
SolomonZelman
  • SolomonZelman
if you are allowed to use the graph, you can get the local max and min from there https://www.desmos.com/calculator/tgu0exzdsv

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SolomonZelman
  • SolomonZelman
for points of inflection we need second derivatiev f''(x).
jmartinez638
  • jmartinez638
F'(x) = x^2 -2x - 35
SolomonZelman
  • SolomonZelman
yes, and the second derivative?
jmartinez638
  • jmartinez638
F''(x) = 2x-2
SolomonZelman
  • SolomonZelman
Yes...
SolomonZelman
  • SolomonZelman
Set f''(x)=0, to solve for inflections but there is one more little test to see if inflections are inflections
SolomonZelman
  • SolomonZelman
say you get x=a, b for f''(x)=0. You need to varify that the concavity is indded changing at x=a and x=b. let's take some number c that is sufficiently close to a (won't use epsilon or any of that), AND find find f''(x-c) and f''(x+c). If they have different signs, then the concavity indeed changed at x-a. (Same for x=b)
SolomonZelman
  • SolomonZelman
Alright, let's go together... f''(x)=0 2x-2=0 x=1
SolomonZelman
  • SolomonZelman
So x=1 is your (only) possible inflection point.
jmartinez638
  • jmartinez638
Ok. So to see if it is an actual inflection point, I am assuming we have to see if the direction of the function actually changes. How do we do this?
SolomonZelman
  • SolomonZelman
Yes, very good.... and for clarifying the direction change, we use the test I mentioned: ---------------------------- If you get x=a for f''(x)=0. You need to varify that the concavity is indeed changing at x=a. let's take some number c that is sufficiently close to a (won't use epsilon or any of that), AND find find f''(x-c) and f''(x+c). If they have different signs, then the concavity indeed changed at x-a. (Same for x=b)
SolomonZelman
  • SolomonZelman
So we can take c=0.05 f''(1-0.05) and f''(1+0.05) if they have different signs then concaavity changes, and x=1 is indded an inflection.
SolomonZelman
  • SolomonZelman
f''(1-0.05)=f''(0.95)=2(0.95)-2 f''(1+0.05)=f''(1.05)=2(1.05)-2
SolomonZelman
  • SolomonZelman
you can tell that 2(0.95) is less than 2, so the first equation is negative. AND you can tell that 2(1.05) is greater than 2, so the second equation is positive.
SolomonZelman
  • SolomonZelman
So, fill in blank; x=1, is indeed the __________ ____\(\bf.\)
SolomonZelman
  • SolomonZelman
if you have some questions, ask
jmartinez638
  • jmartinez638
X = 1 is indeed the inflection point.
jmartinez638
  • jmartinez638
Because there is a sign change on either side of x = 1
jmartinez638
  • jmartinez638
That makes a lot of sense, thanks for the help.
SolomonZelman
  • SolomonZelman
Yw

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