What is the range of f(x)=y^2+8 ?

The range of f(x)=y^2+8 is f(x)>= 8

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range of f(x)=y^2+8

range

f (x) = y^{2} + 8

f (x) \geq 8

Interval Notation

[8, \infty)

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

y^{2} + 8

Inverse of

y^{2} + 8 : y - 8, - y - 8

y - 8, - y - 8

Find the domain of each inverse function

Domain of

y - 8 : y \geq 8

Domain of

- y - 8 : y \geq 8

Combine the intervals

f (x) \geq 8

4 x - 3 y = - 24

\frac{x ^{2} - 25}{\frac{x ^{2}}{x ^{2} - 1}}

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

y = x^{2} - 2 x

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