Horizontal inflection point of a parametric polynomial function

  • #1
greg_rack
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Homework Statement
Given ##y=ax^3+bx^2+2x-3##, find the values of ##a## and ##b## for which the function has an horizontal inflection point at ##x=-1##.
Relevant Equations
none
For ##x=-1## to be an *horizontal* inflection point, the first derivative ##y'## in ##-1## must be zero; and this gives the first condition: ##a=\frac{2}{3}b##.

Now, I believe I should "use" the second derivative to obtain the second condition to solve the two-variables-system, but how?
Since it is an inflection point, shouldn't even the second derivative be zero? But the fact that it's an horizontal one is confusing me.
 
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  • #2
greg_rack said:
For ##x=-1## to be an *horizontal* inflection point, the first derivative ##y'## in ##-1## must be zero; and this gives the first condition: ##a=\frac{2}{3}b##.
Check this :wink:
greg_rack said:
Since it is an inflection point, shouldn't even the second derivative be zero?
Your thinking is correct!
 
  • #3
etotheipi said:
Your thinking is correct!
But then, why does ##y''(-1)=0 \rightarrow b=0##, if the correct result should be ##b=2##?
 
  • #4
greg_rack said:
But then, why does ##y''(-1)=0 \rightarrow b=0##, if the correct result should be ##b=2##?

I think you made an error in working out the second derivative. Shouldn't it be, ##y'(-1) = 3a(-1)^2 + 2b(-1) + 2 = 3a - 2b + 2 \overset{!}{=} 0 \implies 3a = 2b -2##, and then ##y''(-1) = 6a(-1) + 2b \overset{!}{=} 0 \implies 3a = b##?
 
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  • #5
etotheipi said:
I think you made an error in working out the second derivative. Shouldn't it be, ##y'(-1) = 3a(-1)^2 + 2b(-1) + 2 = 3a - 2b + 2 \overset{!}{=} 0 \implies 3a = 2b -2##, and then ##y''(-1) = 6a(-1) + 2b \overset{!}{=} 0 \implies 3a = b##?
Yupp, here's the deal... silly me!
Thanks a lot :)
 
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  • #6
greg_rack said:
Yupp, here's the deal... silly me!
Thanks a lot :)

Don't worry, I do stuff like that all the time :smile:
 
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