What is the f(0)=(t(t-3))/(t-2)+(4-t)/(t-2) ?

The f(0)=(t(t-3))/(t-2)+(4-t)/(t-2) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

f(0)=(t(t-3))/(t-2)+(4-t)/(t-2)

f (0) = \frac{t ( t - 3 )}{t - 2} + \frac{4 - t}{t - 2}

Domain : t < 2 or t > 2

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

Y Intercepts : (0, - 2)

Asymptotes : Slant y = t - 2

Extreme Points : None

Interval Notation

Domain : (- \infty, 2) \cup (2, \infty)

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

Solution steps

Domain of

\frac{t ( t - 3 )}{t - 2} + \frac{4 - t}{t - 2} : t < 2 or t > 2

Range of

\frac{t ( t - 3 )}{t - 2} + \frac{4 - t}{t - 2} : f (t) < 0 or f (t) > 0

Axis interception points of

\frac{t ( t - 3 )}{t - 2} + \frac{4 - t}{t - 2} :

Y Intercepts

: (0, - 2)

Asymptotes of

\frac{t ( t - 3 )}{t - 2} + \frac{4 - t}{t - 2} :

Slant

: y = t - 2

Extreme Points of

\frac{t ( t - 3 )}{t - 2} + \frac{4 - t}{t - 2} :

None

f (x) = x^{4} - 5 x^{3} + 5 x^{2} + 5 x - 6

g (x) = 4 x + 3

h (x) = (x + 1)^{2}

f (x) = 7 + x

f (x) = 8 x^{3} - 2

What is the f(0)=(t(t-3))/(t-2)+(4-t)/(t-2) ?
The f(0)=(t(t-3))/(t-2)+(4-t)/(t-2) is