For what values of the constant c is the function f(x) =cx − 1/x^2+1 increasing∀x. (Hint: don’t try to solve the inequality)

Answer 1

For a function to be increasing, f'(x) > 0 for all x.

Since we are given that f(x) = cx - x^-2 + 1, then f'(x) = c + 2x^-3 > 0 for all x.

Since -∞ < x < ∞, then -∞ < 2 / x³ < ∞ , and so there is no value of c that will make f'(x) > 0 for all x.

However, if f(x) = cx - (x² + 1)^-1, then we have a different situation. Here,

f'(x) = c + 2x / (x² + 1)² > 0 or c > -2x / (x² + 1)² for all x. All we need to do is to determine the minimum of g(x) = -2x / (x² + 1)². To do that, we set g'(r) to 0 and solve for r, then we have c > g(r).

Use Logarithmic Differentiation as follows:

ln(g(x)) = ln(-2) + ln(x) - 2 ln(x² + 1) and differentiate:

g' / g = 0 + 1/x - 4x / (x² + 1) = 0 when (x² + 1) = 4x², which happens when x² = 1/3. Hence r = ±√(1/3).

g(+√(1/3) = -2 √(1/3) / (1/3 +1)² = (- 2/√3) (16/9) = - 32 / (9√3) ≈- 6.1584...

Notice that g(x) is an odd function: g(x) = -g(-x). Hence, g(- √(1/3)) = 6.1584...

Hence any c > 32 / (9√3) will insure that f(x) is always increasing,

Notice that there are some holes in this analysis. For example, ln(-2) is not defined in Reals, but is defind in the Complex Realm, where ln(-2) is a complex constant whose derivative is still zero. Can you find the other hole?