Find the positive x-value where the function f(x)= 1/(3x^2+4) has an inflection point.?

Question

Calculus... I'm following the steps, just missing something somewhere along the line... a lil help?

Answer 1

By the chain rule and power rule, f'(x) = -6x / (3x^2+4)^2

By the quotient rule,

f''(x) = [(3x^2+4)^2 (-6) - (-6x)(2)(3x^2+4)(6x)] / (3x^2+4)^4.

To find the inflection point, you can find where f''(x)=0. For that, you only need to look at the numerator, which needs to be simplified.

For the numerator, you have

-6(3x^2+4)^2 + 72x^2(3x^2+4).

3x^2+4 is always positive, so factor it out. You're left with solving

-6(3x^2+4) + 72x^2 = 0.

Factor out a 6:

-(3x^2+4) + 12x^2 = 0.

Simplify:

9x^2 - 4 = 0.

Now solve 9x^2-4=0. You get x=2/3 and x=-2/3, so 2/3 is the positive value.

NOTE: one of the answers down below INCORRECTLY states that an inflection point occurs when f'(x)=0 AND f''(x)=0. It is not required that f'(x)=0 to have an inflection point.

NOTE 2: In this case, as well as the other one you asked about, checking explicitly for a sign change in the second derivative is not necessary. The second derivative has the form g(x)p(x), where g(x)>0 and p(x) is a second-degree polynomial, which always changes sign at its roots except at a double root, which isn't the case here because the roots are distinct.

Answer 2

What having a horizontal tangent and a factor of inflection propose is that for some fee of the consistent b, the 1st by-product will equivalent 0, and the 2d will replace signs and indications at that factor, which would be the two 0 or undefined. the 1st by-product is y'=4x^3 + b + 8 2d is y''=12x^2 So, if y''=0 at x=0, then y'=0 at x=0, if x=0, then b+8=0, so b= - 8.

Answer 3

The answers of James L and Jim M are basically
correct, but once you solve f"(x) = 0, you must
check if the second derivative changes sign as
the curve passes through each of the points you found.

Answer 4

for inflexion f'(x) = f''(x) = 0

f'(x) = -6x/(3x^2+4)^2
f''(x) = 72x^2/(3x^2+4)^3 - 6/(3x^2+4)^2

f'(x) = 0 when x = 0
when x = 0, f''(x) = -3/8

so the function has NO point of inflection, only a maxima.

If you don't believe me! go to the following address and plot the function to see yourself:
http://www.teachers.ash.org.au/mikemath/calculators.html

Answer 5

Unless I'm missing something
f'(x)=-6x/(9x^4+24x^2+16
the only value for x that gives f'=0 is x=0. This is a maximum not an inflexion pt & is not >0.

Answer 6

calculate f''.

set f'' = 0

solve for x

if those are the steps you're following, then you're doing fine.