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

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

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

range

f (x) = x^{2} + 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} + x - 6 : x \leq - 3 or x \geq 2

Extreme Points of

x^{2} + 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

f (x) = \frac{3}{2 x}

3 x + 2 y = - 6

5 x - 2 y = 10

2 x + 5 y = - 3

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

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