Find the absolute maximum and absolute minimum values of f on the given interval.

Question

f(x) = x/x^2 + 1,    [0, 4]

Answer 1

Let f be the function defined on [0, 4] by

f(x) = x / (x² + 1)

Using the Quotient Rule for derivatives, the derivative of the function f is given by:

f'(x) = [1 * (x² + 1) - x * 2x] / (x² + 1)²

= (x² + 1 - 2x²) / (x² + 1)²

= (-x² + 1) / (x² + 1)²

f'(x) = 0

⇒ (-x² + 1) / (x² + 1)² = 0

⇒ -x² + 1= 0

⇒ (1 - x)(1 + x) = 0

⇒ 1 - x = 0 or 1 + x = 0

⇒ x = 1 or x = -1

So the extrema occur either at x = 1 or at the endpoints of the interval f is defined in.

f(0) = 0 / (0² + 1) = 0

f(1) = 1 / (1² + 1) = 1 / (1 + 1) = 1 / 2

f(4) = 4 / (4² + 1) = 4 / (16 + 1) = 4 / 17

Therefore,

the absolute minimum of f over the interval [0, 4] is 0 and happens at x = 0

the absolute maximum of f over the interval [0, 4] is 1 / 2 and happens at x = 1