find all possible values of f(x)?

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

1 Answer
Jun 16, 2018

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

Explanation:

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]}