What is the range of (0.052x)/(0.9+0.048x) ?

The range of (0.052x)/(0.9+0.048x) is f(x) 13/12

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range of (0.052x)/(0.9+0.048x)

range

\frac{0.052 x}{0.9 + 0.048 x}

f (x) < \frac{13}{12} or f (x) > \frac{13}{12}

Interval Notation

(- \infty, \frac{13}{12}) \cup (\frac{13}{12}, \infty)

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{0.052 x}{0.9 + 0.048 x}

Inverse of

\frac{0.052 x}{0.9 + 0.048 x} : - \frac{0.9 x}{0.048 x - 0.052}

- \frac{0.9 x}{0.048 x - 0.052}

Find the domain of each inverse function

Domain of

- \frac{0.9 x}{0.048 x - 0.052} : x < \frac{13}{12} or x > \frac{13}{12}

Combine the intervals

f (x) < \frac{13}{12} or f (x) > \frac{13}{12}

5 x - 2 y = 3

f (x) = \frac{x ^{2} + x - 6}{x - 4}

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

5 - x + 25

f (x) = - 2 x^{2} + 8 x + 7

What is the range of (0.052x)/(0.9+0.048x) ?
The range of (0.052x)/(0.9+0.048x) is f(x)< 13/12 \lor f(x)> 13/12