Consider the function f(x)=1/(x^6+1)^2.

Question

Consider the function f(x)=1/(x^6+1)^2.

Give the exact x value of a non-stationary inflection point (an inflection point which is not a critical point)

Answer 1

f(x) = (x⁶+1)^-2

f'(x) = -2(x⁶+1)^-3(6x⁵) = -12x⁵(x⁶+1)^-3

f"(x) = -60x⁴(x⁶+1)^-3 + 36x⁵(x⁶+1)^-4(6x⁵) = 12x⁴(x⁶+1)^-4(-5(x⁶+1) + 18x⁶) = 12x⁴(x⁶+1)^-4(13x⁶ - 5)

f"(x) = 0 when x = 0 (also a critical point) or when x = ±(5/13)^1/6

Answer 2

Rewrite f(x) as f(x) = (x⁶ + 1)^-2.

f'(x) = -2(x⁶ + 1)^-3(6x⁵) = -12x⁵(x⁶ + 1)^-3

Make note that f'(x) only equals zero when x = 0 (0 is the only critical number)..

Now find f''(x) using the product rule:

f''(x) = -12x⁵ (-3)(x⁶ + 1)^-4(6x⁵) + (x⁶ + 1)^-3(-60x⁴)

Rewrite f''(x) with positive exponents, simplifying, and setting equal to zero:

216x¹⁰/(x⁶ + 1)⁴ - 60x⁴/(x⁶ + 1)³ = 0

Multiply every term by (x⁶ + 1)⁴ to clear the equation of fractions. Make note that this expression can never equal zero.

216x¹⁰ - 60x⁴(x⁶ + 1) = 0

216x¹⁰ -60x¹⁰ - 60x⁴ = 0

156x¹⁰ - 60x⁴ = 0

Factor out the GCF of 12x⁴:

12x⁴(13x⁶ - 5) = 0

Set each factor equal to zero (zero product property) to determine the hypercritical values (OK, no one uses that term any more--how about the possible x-coordinates of the inflection points).

Obviously one of the values is 0, which is also the x-coordinate of the only critical point.

But 13x⁶ - 5 = 0

13x⁶=5

x⁶ = 5/13

x = ±(5/13)^1/6 (or plus or minus the sixth root of 5/13).

Still have to show that the 2nd derivative changes sign on either side of these hypercritical numbers to prove these are x-coordinates of inflection points.

Check it out here:

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