What is the f(x)=log_{4}(3x+1)+log_{4}(x^2+2) ?

The f(x)=log_{4}(3x+1)+log_{4}(x^2+2) is

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f(x)=log_{4}(3x+1)+log_{4}(x^2+2)

f (x) = lo g_{4} (3 x + 1) + lo g_{4} (x^{2} + 2)

Domain : x > - \frac{1}{3}

Range : - \infty < f (x) < \infty

X Intercepts : (- 0.16901 \dots, 0), Y Intercepts : (0, \frac{1}{2})

Asymptotes : Vertical x = - \frac{1}{3}

Extreme Points : None

Interval Notation

Domain : (- \frac{1}{3}, \infty)

Range : (- \infty, \infty)

Solution steps

Domain of

lo g_{4} (3 x + 1) + lo g_{4} (x^{2} + 2) : x > - \frac{1}{3}

Range of

lo g_{4} (3 x + 1) + lo g_{4} (x^{2} + 2) : - \infty < f (x) < \infty

Axis interception points of

lo g_{4} (3 x + 1) + lo g_{4} (x^{2} + 2) :

X Intercepts

: (- 0.16901 \dots, 0),

Y Intercepts

: (0, \frac{1}{2})

Asymptotes of

lo g_{4} (3 x + 1) + lo g_{4} (x^{2} + 2) :

Vertical

: x = - \frac{1}{3}

Extreme Points of

lo g_{4} (3 x + 1) + lo g_{4} (x^{2} + 2) :

None

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

f (x) = 2 x^{2} - 3 x - 7

f (x) = 2 x^{2} - 3 x + 9

y = tan (x + \frac{π}{2})

f (1) = 3 x - 1

What is the f(x)=log_{4}(3x+1)+log_{4}(x^2+2) ?
The f(x)=log_{4}(3x+1)+log_{4}(x^2+2) is