What is the range of f(x)= 1/(sqrt(x^2+2x+3)) ?

The range of f(x)= 1/(sqrt(x^2+2x+3)) is 0<f(x)<= 1/(sqrt(2))

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range of f(x)= 1/(sqrt(x^2+2x+3))

range

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

0 < f (x) \leq \frac{1}{2}

Interval Notation

(0, \frac{1}{2}]

Solution steps

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

Domain of

\frac{1}{x ^{2} + 2 x + 3} : - \infty < x < \infty

Extreme Points of

\frac{1}{x ^{2} + 2 x + 3} :

Maximum

(- 1, \frac{1}{2})

Find the range for the interval

- \infty < x < \infty : 0 < f (x) \leq \frac{1}{2}

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

0 < f (x) \leq \frac{1}{2}

f (x) = (x - 3) x

x^{2} + 1

- 3 + ln (x)

f (x) = 1 - e^{- 0.01 x}

What is the range of f(x)= 1/(sqrt(x^2+2x+3)) ?
The range of f(x)= 1/(sqrt(x^2+2x+3)) is 0<f(x)<= 1/(sqrt(2))