What is the range of 2ln(x^2+1) ?

The range of 2ln(x^2+1) is f(x)>= 0

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range of 2ln(x^2+1)

range

2 ln (x^{2} + 1)

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

2 ln (x^{2} + 1)

Inverse of

2 ln (x^{2} + 1) : e^{\frac{x}{2}} - 1, - e^{\frac{x}{2}} - 1

e^{\frac{x}{2}} - 1, - e^{\frac{x}{2}} - 1

Find the domain of each inverse function

Domain of

e^{\frac{x}{2}} - 1 : x \geq 0

Domain of

- e^{\frac{x}{2}} - 1 : x \geq 0

Combine the intervals

f (x) \geq 0

(5) (1.4)

m = - 2, (- 3, 14)

f (x) = \frac{1}{1 + x ^{2}} - 3 x^{2}

y = \frac{( x ^{2} )}{x ^{2} - 16}

f (x) = \frac{2 x + 3}{5 x + 4}