What is the f(x)=1+x^2*(1+x^2)^{1/2} ?

The f(x)=1+x^2*(1+x^2)^{1/2} is

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f(x)=1+x^2*(1+x^2)^{1/2}

f (x) = 1 + x^{2} \cdot (1 + x^{2})^{\frac{1}{2}}

Domain : - \infty < x < \infty

Range : f (x) \geq 1

Y Intercepts : (0, 1)

Asymptotes : None

Extreme Points : Minimum (0, 1)

Interval Notation

Domain : (- \infty, \infty)

Range : [1, \infty)

Solution steps

Domain of

1 + x^{2} \cdot (1 + x^{2})^{\frac{1}{2}} : - \infty < x < \infty

Range of

1 + x^{2} \cdot (1 + x^{2})^{\frac{1}{2}} : f (x) \geq 1

Axis interception points of

1 + x^{2} \cdot (1 + x^{2})^{\frac{1}{2}} :

Y Intercepts

: (0, 1)

Asymptotes of

1 + x^{2} \cdot (1 + x^{2})^{\frac{1}{2}} :

None

Extreme Points of

1 + x^{2} \cdot (1 + x^{2})^{\frac{1}{2}} :

Minimum

(0, 1)

y = x + lo g_{3} (x) - cos (x)

p (x) = x^{2} - 4

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

f (x) = \frac{lo g _{10} ( x )}{1 + x}

g (x) = \frac{1}{80} \cdot x^{2} - \frac{1}{2} \cdot x + 20