What is the range of 1/(sqrt(x)+19) ?

The range of 1/(sqrt(x)+19) is 0<f(x)<= 1/19

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range of 1/(sqrt(x)+19)

range

\frac{1}{x + 19}

0 < f (x) \leq \frac{1}{19}

Interval Notation

(0, \frac{1}{19}]

Solution steps

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

Domain of

\frac{1}{x + 19} : x \geq 0

Extreme Points of

\frac{1}{x + 19} :

Maximum

(0, \frac{1}{19})

Find the range for the interval

0 \leq x < \infty : 0 < f (x) \leq \frac{1}{19}

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

0 < f (x) \leq \frac{1}{19}

f (x) = \frac{1}{2}

f (x) = - x^{2} - 3 x

\frac{x + 4}{x + 1}

2 x - 3 y = 9

y = x^{2} + 2 x + 1

What is the range of 1/(sqrt(x)+19) ?
The range of 1/(sqrt(x)+19) is 0<f(x)<= 1/19