How do you find points of inflection of a derivative function?

Question

f(x)=2x^3 -10x +3 you find the second derivative to determine if it is concave up or down, but I don't know how to find points of inflection. And how would you find the equation of all vertical and horizontal asymptotes?

Answer 1

points of inflection are the places where the second derivative of a function equals zero.
So in this case f'(x)=6x^2-10, and f''(x)=12x. Now if you set f''(x)=0 it is the same as 12x=0 and dividing both sides by 12 you have that x=0. Thus your inflection point is at x=0.

Now recall that f(x)=2x^3 -10x +3 is a third degree polynomial, which looks some what like and s curve, and it does indeed have only one place where the concavity changes from up to down, so having only one inflection point makes sense.

Polynomials do not have vertical asymptotes nor do they have horizontal ones. You might be thinking about rational functions, which are polynomials divided by polynomials.

If that is indeed what you are refering to them, to find the vercital asymptotes you set the denominator of the function equal to zero and solve for x. These are then the vertical asymptotes.

For horizontal asymptotes it is more complicated I suggest you look it up in a math book, the simplst case is like this:

If the highest degrees on the numerator and the denoinator are equal then the ratio of the coefficients is the horizontal asymptote. Example f(x)=(2x^3 -10x +3)/(3x^3-17), the highest degree of the umerator is 3 which is the same as the highest degree of the denominator, so the horizontal asymptote is y=2/3.

Hope this helps!

Answer 2

This Site Might Help You.

RE:
How do you find points of inflection of a derivative function?
f(x)=2x^3 -10x +3 you find the second derivative to determine if it is concave up or down, but I don't know how to find points of inflection. And how would you find the equation of all vertical and horizontal asymptotes?

Answer 3

Let a = 4, so a' = 0 Let b = 9 + x^2, so b' = 2x Since y = a/b, use the Quotient Rule: y' = (a'b - ab') / (b^2) y' = ((0)(9 + x^2) - (4)(2x)) / ((9 + x^2)^2) y' = (-8x) / ((9 + x^2)^2) Let c = -8x, so c' = -8 Let d = 81 + 18x^2 + x^4, so d' = 36x + 4x^3 Since y' = c/d, use the Quotient Rule: y'' = (c'd - cd') / (d^2) y'' = ((-8)(81 + 18x^2 + x^4) - (-8x)(36x + 4x^3)) / (((9 + x^2)^2)^2) y'' = (-648 - 144x^2 - 8x^4 + 288x^2 + 32x^4) / ((9 + x^2)^4) y'' = (-648 + 144x^2 + 24x^4) / ((9 + x^2)^4) Factor the numerator: y'' = (24(9 + x^2)(x^2 - 3)) / ((9 + x^2)^4) Cancel out 9 + x^2: y'' = (24(x^2 - 3)) / ((9 + x^2)^3) To find the points of inflection, we need the numerator to be zero: 24(x^2 - 3) = 0 x^2 - 3 = 0 x^2 = 3 x = +/- sqrt(3)

Answer 4

In order to find Vertical Asymptotes you have to set the denominator equal to zero

Now Horizontal Asymptotes

Consider This Function: F(x) = Ax^n / Bx^m (A & B Are Coefficients) & (N & M are powers or otherswise known as a degree)

Now if n < m then you have a horizontal asymptote that is y = 0 this works everytime.

If n = m then the horizontal asymtote is the two coefficient and in this case they are A / B

If n > m there is no horizontal asymptote and you can more to finding the oblique asymptote.

Answer 5

Points of inflection occur when the 2nd derivative is equal to zero.

Answer 6

do you not subtract something at this point , but we need the equation...

I see lens here!

Just messing... The first is correct... baffled the hell out of me though!