QUESTIONS - Lillian.Blackwell
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Lillian.Blackwell to me
Could you please help me with finding all the points of inflection with the
below problem:
F(x) =x^4-18x^2+4
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For points of inflection the following criteria must be satisfied……
AT x = a
1.F”(a)=0 or does not exist
2. But F’(a) exists and
3.F”(x) changes sign when passing through the point x=a
Then the point [x=a and y=F(a)]
is the point of inflection of the curve.
F(x) =x^4-18x^2+4
F’(x) = 4x^3-36x
F”(x) =12x^2-36
F”(x)=0 at 12(x^2-3)=0…or…x = =sqrt(3)….or…..- sqrt(3)
F’(x) exists at x=sqrt(3)…..&……x= -sqrt(3)
Now make a table as follows to study change of sign in F(x)
x…x is less than - sqrt(3)………-sqrt(3)
sign of F”………+ ve……………………- ve …...........………………+ve
since F”(x) changes sign while passing through x=-sqrt(3)….and ….x=sqrt(3)
There are 2 points of inflection..they are
[-sqrt(3),9-54+4]……and ……[sqrt(3),9-54+4]
that is [-sqrt(3),-41] and [sqrt(3),-41]
posted by mathsmanagementcommonsense at 10:03 AM
1 Comments:
Need some help!!
Consider a second order differential equation of the form y" = f(y,y'). If we let v equal y', we obtain v' = f(y,v). This equation uses the variables x, y and v and hence is not the form of the first order equations. However, it is possible to eliminate the variable x by thinking of y as an independent variable; the chain rule gives
dv/dx = (dv/dy)(dy/dx) = (dv/dy)v,
and hence the original differential equation can be written as
v (dv/dy) = f(y,v)
Provided this equation can be solved, we obtain v as a function of y. A relation between y and x results from solving
dy/dx = v(y)
Again there are two arbitary constants in the final result. Use this technique to solve the following differential equation.
y" + y(y')^3 = 0
thanks
9:29 AM
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