What is the range of f(x)=(x-7)/(x^2+7)+1 ?

The range of f(x)=(x-7)/(x^2+7)+1 is -sqrt(2/7)+1/2 <= f(x)<1\lor 1<f(x)<= sqrt(2/7)+1/2

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range of f(x)=(x-7)/(x^2+7)+1

range

f (x) = \frac{x - 7}{x ^{2} + 7} + 1

- \frac{2}{7} + \frac{1}{2} \leq f (x) < 1 or 1 < f (x) \leq \frac{2}{7} + \frac{1}{2}

Interval Notation

[- \frac{2}{7} + \frac{1}{2}, 1) \cup (1, \frac{2}{7} + \frac{1}{2}]

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{x - 7}{x ^{2} + 7} + 1

Inverse of

\frac{x - 7}{x ^{2} + 7} + 1 : \frac{1 + 1 - 28 x ( x - 1 )}{2 ( x - 1 )}, \frac{1 - 1 - 28 x ( x - 1 )}{2 ( x - 1 )}

\frac{1 + 1 - 28 x ( x - 1 )}{2 ( x - 1 )}, \frac{1 - 1 - 28 x ( x - 1 )}{2 ( x - 1 )}

Find the domain of each inverse function

Domain of

\frac{1 + 1 - 28 x ( x - 1 )}{2 ( x - 1 )} : - \frac{2}{7} + \frac{1}{2} \leq x < 1 or 1 < x \leq \frac{2}{7} + \frac{1}{2}

Domain of

\frac{1 - 1 - 28 x ( x - 1 )}{2 ( x - 1 )} : - \frac{2}{7} + \frac{1}{2} \leq x < 1 or 1 < x \leq \frac{2}{7} + \frac{1}{2}

Combine the intervals

- \frac{2}{7} + \frac{1}{2} \leq f (x) < 1 or 1 < f (x) \leq \frac{2}{7} + \frac{1}{2}

\frac{x ^{2} + 3 x}{x ^{2} - 2 x - 15}

f (x) = 3 x - 1

\frac{x ^{2} + 5 x}{x ^{2} + 7 x + 10}

f (x) = ln (x) + 5

f (x) = e^{2.5 x^{2}}

What is the range of f(x)=(x-7)/(x^2+7)+1 ?
The range of f(x)=(x-7)/(x^2+7)+1 is -sqrt(2/7)+1/2 <= f(x)<1\lor 1<f(x)<= sqrt(2/7)+1/2