Question 1030349: Let d,e and g be real numbers and consider the function F(x) = (x-d)(x-e)(x-g). Using Calculus, find the x-coordinate of the point of the inflection for the given function.
Found 2 solutions by mathmate, richard1234:
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
Let d,e and g be real numbers and consider the function F(x) = (x-d)(x-e)(x-g). Using Calculus, find the x-coordinate of the point of the inflection for the given function.
Solution:
Given a function F(x) differentiable around x=x0 where x0 locates a point of inflection.
A point of inflection located at x=x0 has the property that F"(x0)=0, which means that this is only a necessary condition.
A sufficient condition is that F"(x) must have different signs on either side of x0.
With the above information, we will search for the point of inflection of the function F(x)=(x-d)(x-e)(x-g).
- Differentiate F(x) twice to get
F'(x)=(x-e)(x-g)+(x-d)(x-g)+(x-d)(x-e), and
F"(x)=6x-2g-2e-2d
- Since F"(x)=0 is a necessary condition for an inflection point, we solve for x in F"(x)=0 to get
x0=(d+e+g)/3.
- We need to check if the point is in fact a point of inflection using the sufficient condition:
- Let ε= a very small number, then x0+ε and x0-ε are on each side of x0.
- We calculate
Q=F"(x0+ε)*F"(x0-ε)
=[+6ε][-6ε]
=-36ε²
<0 (less than zero, therefore satisfies the sufficient condition)
- Hence x0=(d+e+g)/3 is an inflection point.
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! F(x) equals x^3 - (d + e + g)x^2 + (de + eg + dg)x - deg.
Then F'(x) = 3x^2 - 2(d + e + g)x + de + eg + deg, and F''(x) = 6x - 2(d + e + g). Points of inflection occur when F''(x) = 0 and F changes concavity.
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