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

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

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

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

Domain : x > \frac{1}{2}

Intercepts : None

Asymptotes : Vertical x = \frac{1}{2}

Extreme Points : Minimum (\frac{4}{5}, 2 + 3 lo g_{2} (\frac{9}{5}) - lo g_{4} (\frac{3}{5}))

Interval Notation

Domain : (\frac{1}{2}, \infty)

Solution steps

Domain of

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

Axis interception points of

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

None

Asymptotes of

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

Vertical

: x = \frac{1}{2}

Extreme Points of

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

Minimum

(\frac{4}{5}, 2 + 3 lo g_{2} (\frac{9}{5}) - lo g_{4} (\frac{3}{5}))

f (n) = (lo g_{10} (n))^{2}

f (x) = cos (3 x) cos (2 x)

f (x) = \frac{2 x ^{2} + 7 x + 3}{x ^{2} + 2 x - 8}

y = ∣ x + 4 ∣ - 2

f (x) = x^{2} - 12 x + 30

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