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Popular >

f(n)=(2^n-1)/(4^n-1)

Solution

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

Solution

Domain : n < 0 or n > 0

Range : f (x) < 0 or f (x) > 0

Intercepts : None

Asymptotes : Horizontal y = 0, y = 1

Extreme Points : None

+1

Interval Notation

Domain : (- \infty, 0) \cup (0, \infty)

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

Solution steps

Domain of

\frac{2 ^{n} - 1}{4 ^{n} - 1} : n < 0 or n > 0

Range of

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

Axis interception points of

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

None

Asymptotes of

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

Horizontal

: y = 0, y = 1

Extreme Points of

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

None

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f(x)=2x^2+5x-17

f (x) = 2 x^{2} + 5 x - 17

Frequently Asked Questions (FAQ)

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