Find Largest Value of Function

Find the largest value of the function f(x) = 1/(x^4 + 2x^2 + 1).

This is soooo simple! The complex calculus-type solutions you received are correct, but common sense is all you need, at least for this problem. If the denominator was more complex, then those techniques would probably be needed.

It's the largest when the denominator is the smallest.
When x=0, denominator = 1
For any other x, denominator is > 1. (Even powers of x are always positive, so it can't get less than 1.)
So largest value of f is f(0) = 1/1 = 1

(x^4 + 2x^2 + 1) can be written as (x^2+1)^2
So, the expression becomes 1/(x^2+1)^2
x^2 is always greater than or equal to 0.
SO, the minimum value of the denominator is 1, when x^2 is 0.
Therefore, the maximum value of the function is 1

f(x) = 1/(x^4 + 2x^2 + 1)

To find the largest value of this function, let us first factor the denominator prior to taking the derivative.

f(x) = 1/[x^2 + 1]^2
f(x) = (x^2 + 1)^(-2), which means
f'(x) = (-2)(x^2 + 1)^(-1) (2x)
f'(x) = (-2)/(x^2 + 1) * (2x)
f'(x) = -4x/(x^2 + 1)

Now we find the critical values by making f'(x) = 0.

0 = -4x/(x^2 + 1)

Critical values are what make f'(x) = 0 or what makes f'(x) is undefined. f'(x) is never going to be undefined, since x^2 + 1 = 0 has no real solutions. That means f'(x) = 0 when the numerator is 0, i.e.
-4x = 0
x = 0

So 0 is the only critical value. Make a number line / sign diagram to determine the intervals this function is increasing/decreasing.

. . . . . . . . (0) . . . . . . . .

Determine the sign of f'(x) by testing a value such that x < 0, and a value such that x > 0.
Let x = -1. Then f'(-1) = -4(-1)/((-1)^2 + 1) = (-4)(-1)/2 = 4, which is posi