What is the range of f(x)=sqrt(x^2-4)+1 ?

The range of f(x)=sqrt(x^2-4)+1 is f(x)>= 1

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range of f(x)=sqrt(x^2-4)+1

range

f (x) = x^{2} - 4 + 1

f (x) \geq 1

Solution steps

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

Domain of

x^{2} - 4 + 1 : x \leq - 2 or x \geq 2

Extreme Points of

x^{2} - 4 + 1 :

Minimum

(- 2, 1),

Minimum

(2, 1)

Find the range for the interval

- \infty < x \leq - 2 : 1 \leq f (x) < \infty

Find the range for the interval

2 \leq x < \infty : 1 \leq f (x) < \infty

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

f (x) \geq 1 or f (x) \geq 1

f (x) \geq 1

f (x) = \frac{5 x - 15}{2 x - 9}

(- 4 - 7), (- 4 - 6)

f (x) = (x - 5)^{2}

f (x) = - - x^{2} - 6 x

f (x) = 4 x^{3} - 33 x^{2} - 36 x + 2

What is the range of f(x)=sqrt(x^2-4)+1 ?
The range of f(x)=sqrt(x^2-4)+1 is f(x)>= 1