Calculus need help please

Question

7. Find the exact x-value where the function  () attains a maximum value. An estimated answer or a calculator answer will not earn any credit.

Answer 1

This looks like a trick question because f(x) never attains a global maximum.

Notice that f(x) -> ∞ as x -> ∞.

But let's go through the motions.

First lets calculate the fist and second derivative:

f'(x) = 1 + 2x/(x²-1)

f"(x) = 2/(x²-1) - 4x²/(x²-1)²

In general, a function attains its maximum value at a critical point, or at a boundary

of its domain.

The domain of our function f(x) is the set of x making the argument on ln() positive:

x² - 1 > 0

x < -1 or x > 1

Thus the domain has no boundary to consider.

A critical point is where the derivative is 0:

1 + 2x/(x²-1) = 0

x² + 2x - 1 = 0 (1)

x = (-2 ± √(4 + 4))/2 = -1±√2

The solution -1+√2 is in the interval (0,1), where f(x) is undefined.

So we have just one critical point, let's call it c = -1-√2.

To see whether it is local minimum or maximum, see whether f"(c) > 0 or f"(c) < 0.

To make the calculation easier, notice that the point c satisfies (1), so

c² - 1 = -2c

So

f"(c) = 2/-2c - 4c²/4c² = -1/c - 1 = -1/(-1-√2) - 1 = -√2/(1+√2) < 0

Therefore the critical point c is a local maximum.

But it is only LOCAL, not GLOBAL.