What is the range of f(x)=2+sqrt(9+x^2) ?

The range of f(x)=2+sqrt(9+x^2) is f(x)>= 5

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range of f(x)=2+sqrt(9+x^2)

range

f (x) = 2 + 9 + x^{2}

f (x) \geq 5

Interval Notation

Solution steps

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

Domain of

2 + 9 + x^{2} : - \infty < x < \infty

Extreme Points of

2 + 9 + x^{2} :

Minimum

(0, 5)

Find the range for the interval

- \infty < x < \infty : 5 \leq f (x) < \infty

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

f (x) \geq 5

\frac{2 x ^{2} + 4 x + 2}{x ^{2} - 1}

4 csc^{4} (x) cot^{6} (x) d x

f (x) = \frac{4 x + 1}{( x + 3 ) ( x - 5 )}

f (x) = \frac{1}{7 4 + x}

5 x + 3 y = 3

What is the range of f(x)=2+sqrt(9+x^2) ?
The range of f(x)=2+sqrt(9+x^2) is f(x)>= 5