What is the f(x)=sqrt(2x+1+2\sqrt{x^2+x-2)} ?

The f(x)=sqrt(2x+1+2\sqrt{x^2+x-2)} is

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

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

Domain : x \geq 1

Range : f (x) \geq 3

Intercepts : None

Asymptotes : None

Extreme Points : Minimum (1, 3)

Interval Notation

Domain : [1, \infty)

Range : [3, \infty)

Solution steps

Domain of

2 x + 1 + 2 x^{2} + x - 2 : x \geq 1

Range of

2 x + 1 + 2 x^{2} + x - 2 : f (x) \geq 3

Axis interception points of

2 x + 1 + 2 x^{2} + x - 2 :

None

Asymptotes of

2 x + 1 + 2 x^{2} + x - 2 :

None

Extreme Points of

2 x + 1 + 2 x^{2} + x - 2 :

Minimum

(1, 3)

y = \frac{ln ( x + 1 )}{x}

y = 3 x^{2} - 6 x - 1

f (x) = 4 x^{3} - 7

f (x) = \frac{( x ^{2} + 4 x + 4 )}{( x + 2 )}

f (x) = sin (x + 5)

What is the f(x)=sqrt(2x+1+2\sqrt{x^2+x-2)} ?
The f(x)=sqrt(2x+1+2\sqrt{x^2+x-2)} is