What is the range of f(x)=sqrt(4x^2+20) ?

The range of f(x)=sqrt(4x^2+20) is f(x)>= 2sqrt(5)

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range of f(x)=sqrt(4x^2+20)

range

f (x) = 4 x^{2} + 20

f (x) \geq 25

Interval Notation

[25, \infty)

Solution steps

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

Domain of

4 x^{2} + 20 : - \infty < x < \infty

Simplify

4 x^{2} + 20 : 2 x^{2} + 5

Extreme Points of

2 x^{2} + 5 :

Minimum

(0, 25)

Find the range for the interval

- \infty < x < \infty : 25 \leq f (x) < \infty

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

f (x) \geq 25

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

6 x - 3 y = 3

lo g_{2} (3 x)

f (x) = \frac{2 x ^{2} - 7 x - 5}{2 x + 3}

f (x) = \frac{( x ^{4} - 1 )}{x}

What is the range of f(x)=sqrt(4x^2+20) ?
The range of f(x)=sqrt(4x^2+20) is f(x)>= 2sqrt(5)