What is the range of f(x)= t/(sqrt(t+1)) ?

The range of f(x)= t/(sqrt(t+1)) is -infinity <f(x)<infinity

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range of f(x)= t/(sqrt(t+1))

range

f (x) = \frac{t}{t + 1}

- \infty < f (x) < \infty

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{t}{t + 1} : t > - 1

Extreme Points of

\frac{t}{t + 1} :

None

Find the range for the interval

- 1 < t < \infty : - \infty < f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

- \infty < f (x) < \infty

f (x) = 4 x^{3} + 21 x^{2} + 18 x + 2

f (x) = 18 x^{\frac{2}{3}} - 6 x

f (x) = \frac{3 x}{7 x - 6}

f (x) = 1 - x

6 tan (0.2 x)

What is the range of f(x)= t/(sqrt(t+1)) ?
The range of f(x)= t/(sqrt(t+1)) is -infinity <f(x)<infinity