What is the f(x)=log_{2}(2x-1/2) ?

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f(x)=log_{2}(2x-1/2)

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

Domain : x > \frac{1}{4}

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

X Intercepts : (\frac{3}{4}, 0)

Asymptotes : Vertical x = \frac{1}{4}

Extreme Points : None

Interval Notation

Domain : (\frac{1}{4}, \infty)

Range : (- \infty, \infty)

Solution steps

Domain of

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

Range of

lo g_{2} (2 x - \frac{1}{2}) : - \infty < f (x) < \infty

Axis interception points of

lo g_{2} (2 x - \frac{1}{2}) :

X Intercepts

: (\frac{3}{4}, 0)

Asymptotes of

lo g_{2} (2 x - \frac{1}{2}) :

Vertical

: x = \frac{1}{4}

Extreme Points of

lo g_{2} (2 x - \frac{1}{2}) :

None

f (x) = 2 x^{2} - 8 x + 16

r (x) = \frac{x ^{2} + 5 x}{25 - x ^{2}}

f (x) = sin^{2} (x) cos^{2} (x) d x

{x + 4 : - 3 \leq x < 1, 9 : x = 1, - x + 3 : x > 1}

f (x) = x^{2} + 2 x - 15