How do you find all relative extrema for #f(x) = 8/(x^2+2)#?

Question

Luke Alvarez · Accepted Answer

See explanation section, below.

#f(x) = 8/(x^2+2)# #f'(x) = (-16x)/(x^2+2)^2# #f'(x)# is never undefined and is #0# only at #x=0#. So the only critical number is #0#. The denominator of #f'# is always positive, so the sign of #f'# is the same as that of #-16x#, which is simply the opposite of the sign of #x#. On #(-oo,0)#, #f'(x) > 0# and on #(0.oo)#, #f'(x) < 0#. Therefore, #f(0) = 4# is a local maximum. There is a local maximum of #4# (at #0#)

Evelyn Agard · Answer

undefined To find the relative extrema for $ f(x) = \frac{8}{x^2 + 2} $, we first find the critical points by taking the derivative and setting it equal to zero. Then, we check the second derivative to determine the nature of the critical points.

1. Find the derivative of $ f(x) $:

$$ f'(x) = \frac{-16x}{(x^2 + 2)^2} $$

2. Set $ f'(x) $ equal to zero and solve for $ x $ to find critical points:

$$ \frac{-16x}{(x^2 + 2)^2} = 0 $$

This equation is satisfied when $ x = 0 $.

3. Find the second derivative of $ f(x) $:

$$ f''(x) = \frac{32x(x^2 - 6)}{(x^2 + 2)^3} $$

4. Evaluate $ f''(0) $ to determine the nature of the critical point at $ x = 0 $:

$$ f''(0) = \frac{32 \cdot 0(0^2 - 6)}{(0^2 + 2)^3} = 0 $$

Since $ f''(0) = 0 $, the second derivative test is inconclusive.

Therefore, the critical point at $ x = 0 $ could be a relative minimum, maximum, or neither.

There are no other critical points.

Thus, the function $ f(x) = \frac{8}{x^2 + 2} $ has a relative extremum at $ x = 0 $, but we cannot determine whether it is a minimum, maximum, or neither using the second derivative test.