find all possible values of f(x)?

Question

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

2417 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-16T14:33:38

#f(x) in (-1, +1/3]#

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

#f(x)# is defined #forall x in RR -># the domain of #f(x)# is #(-oo,+oo)#

Now consider #lim_(x->+-oo) f(x)#

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

#lim_(x->+oo) f(x) = (0-1)/(1+0) = -1#

#lim_(x->-oo) f(x) = (0-1)/(1+0) = -1#

Thus, #f(x)_min -> -1#

Now let's find #f(x)_max#

The maximum value of #f(x)# will occur for the value of x where the numerator is a maximum and the denominator is a minimum,

#(1-x^2)_max -> x=0# and #(x^2+3)_min -> x=0#

#-> f(x)_max = f(0) = 1/3#

Thus the range (and hence all possible values) of #f(x)# is #(-1,+1/3]#

We can observe this result from the graph of #f(x)# below.

graph{(1-x^2)/(x^2+3) [-5.54, 5.556, -2.783, 2.764]}