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Popular >

f(x)=(|4x^2-1|)/(2x-1)

Solution

f (x) = \frac{4 x ^{2} - 1}{2 x - 1}

Solution

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

Range : f (x) \leq 0 or f (x) > 2

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

Asymptotes : Slant y = 2 x + 1

Extreme Points : Maximum (- \frac{1}{2}, 0)

+1

Interval Notation

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

Range : (- \infty, 0] \cup (2, \infty)

Solution steps

Domain of

\frac{4 x ^{2} - 1}{2 x - 1} : x < \frac{1}{2} or x > \frac{1}{2}

Range of

\frac{4 x ^{2} - 1}{2 x - 1} : f (x) \leq 0 or f (x) > 2

Axis interception points of

\frac{4 x ^{2} - 1}{2 x - 1} :

X Intercepts

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

Y Intercepts

: (0, - 1)

Asymptotes of

\frac{4 x ^{2} - 1}{2 x - 1} :

Slant

: y = 2 x + 1

Extreme Points of

\frac{4 x ^{2} - 1}{2 x - 1} :

Maximum

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

Popular Examples

y=[tan(x^2+1)]^4

y = [tan (x^{2} + 1)]^{4}

p(x)= 3/2 x^3+2x^2+1/2 x+1

p (x) = \frac{3}{2} x^{3} + 2 x^{2} + \frac{1}{2} x + 1

f(x)=(ln(1+x))/((1+x^2))

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

f(x)=2e^{x^2}-5x

f (x) = 2 e^{x^{2}} - 5 x

y = \frac{1}{2} x^{2} - 2 x

Frequently Asked Questions (FAQ)

What is the f(x)=(|4x^2-1|)/(2x-1) ?
The f(x)=(|4x^2-1|)/(2x-1) is