What is the range of f(x)=sqrt(x^2+5x+6) ?

The range of f(x)=sqrt(x^2+5x+6) is f(x)>= 0

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range of f(x)=sqrt(x^2+5x+6)

range

f (x) = x^{2} + 5 x + 6

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

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

Domain of

x^{2} + 5 x + 6 : x \leq - 3 or x \geq - 2

Extreme Points of

x^{2} + 5 x + 6 :

Minimum

(- 3, 0),

Minimum

(- 2, 0)

Find the range for the interval

- \infty < x \leq - 3 : 0 \leq f (x) < \infty

Find the range for the interval

- 2 \leq x < \infty : 0 \leq f (x) < \infty

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

f (x) \geq 0 or f (x) \geq 0

f (x) \geq 0

- \frac{1}{x} - 2

\frac{1}{3 - 5 x}

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

f (x) = - 3 - x - 2

f (x) = \frac{( x - 4 )}{3 x - x ^{2}}

What is the range of f(x)=sqrt(x^2+5x+6) ?
The range of f(x)=sqrt(x^2+5x+6) is f(x)>= 0