What interval is #F(x) = (x^2)/(x^2+3)# increasing, decreasing?

Question

Luke Alvarez · Accepted Answer

#F(x)# is increasing for #x in (0, +oo)# and decreasing for #x in (-oo, 0)#

#F(x) = x^2/(x^2+3)#

#= 1/(1+3/x^2)#

Hence: #Lim_"x->+oo" F(x) = 1# and #Lim_"x->-oo" F(x) = 1#

Also notice, #F(0) = 0# which is an absolute minimum for #F(x)#

Therefore: #F(x)# decreases from 1 for #x<0# and increases to 1 for #x>0#

This can be seen from the graph of #F(x)# below:

graph{x^2/(x^2+3) [-7.025, 7.02, -3.51, 3.51]}

Evelyn Agard · Answer

undefined To determine the intervals where $ F(x) = \frac{x^2}{x^2 + 3} $ is increasing or decreasing, we need to analyze the behavior of its derivative $ F'(x) $.

Let's find $ F'(x) $ first:

$ F(x) = \frac{x^2}{x^2 + 3} $

Using the quotient rule, we get:

$ F'(x) = \frac{(2x)(x^2 + 3) - (x^2)(2x)}{(x^2 + 3)^2} $

$ F'(x) = \frac{2x(x^2 + 3) - 2x^3}{(x^2 + 3)^2} $

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

$ F'(x) = \frac{6x}{(x^2 + 3)^2} $

To find the intervals of increase and decrease, we need to find the critical points by setting $ F'(x) $ equal to zero:

$ \frac{6x}{(x^2 + 3)^2} = 0 $

This equation has no real solutions because the numerator $ 6x $ can never be zero.

Now, we need to determine the intervals where $ F'(x) $ is positive and where it's negative.

Since the numerator $ 6x $ is always positive for $ x \neq 0 $, $ F'(x) $ will always be positive.

Therefore, $ F(x) = \frac{x^2}{x^2 + 3} $ is increasing for all real values of $ x $. There are no intervals of decrease.