What is the range of sqrt(x^2+8x+14) ?

The range of sqrt(x^2+8x+14) is f(x)>= 0

Popular >

range of sqrt(x^2+8x+14)

range

x^{2} + 8 x + 14

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} + 8 x + 14 : x \leq - 2 - 4 or x \geq 2 - 4

Extreme Points of

x^{2} + 8 x + 14 :

Minimum

(- 4 - 2, 0),

Minimum

(- 4 + 2, 0)

Find the range for the interval

- \infty < x \leq - 2 - 4 : 0 \leq f (x) < \infty

Find the range for the interval

2 - 4 \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 (t) = ln (t + 1)

f (x) = x^{2}, - 2 \leq x \leq 5

4 x^{2} - 4 x + 9

f (x) = - (5)^{x} + 5

(- \frac{1}{3}, \frac{1}{5}), (- \frac{11}{2}, \frac{9}{10})