What is the range of (3x^2-18x+24)/(x^2-4x) ?

The range of (3x^2-18x+24)/(x^2-4x) is f(x) 3

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range of (3x^2-18x+24)/(x^2-4x)

range

\frac{3 x ^{2} - 18 x + 24}{x ^{2} - 4 x}

f (x) < \frac{3}{2} or \frac{3}{2} < f (x) < 3 or f (x) > 3

Interval Notation

(- \infty, \frac{3}{2}) \cup (\frac{3}{2}, 3) \cup (3, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{3 x ^{2} - 18 x + 24}{x ^{2} - 4 x} : x < 0 or 0 < x < 4 or x > 4

Simplify

\frac{3 x ^{2} - 18 x + 24}{x ^{2} - 4 x} : \frac{3 ( x - 2 )}{x}

Extreme Points of

\frac{3 ( x - 2 )}{x} :

None

Find the range for the interval

- \infty < x < 0 : 3 < f (x) < \infty

Find the range for the interval

0 < x < 4 : - \infty < f (x) < \frac{3}{2}

Find the range for the interval

4 < x < \infty : \frac{3}{2} < f (x) < 3

Combine the ranges of all domain intervals to obtain the function range

f (x) > 3 or f (x) < \frac{3}{2} or \frac{3}{2} < f (x) < 3

f (x) < \frac{3}{2} or \frac{3}{2} < f (x) < 3 or f (x) > 3

f (x) = 4 cos (x - \frac{π}{3}) + 2

\frac{x + 3}{x ^{2} - 4}

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

f (x) = x^{2} + \frac{( 26 - 2 x ) ^{2}}{9}

f (x) = 1

What is the range of (3x^2-18x+24)/(x^2-4x) ?
The range of (3x^2-18x+24)/(x^2-4x) is f(x)< 3/2 \lor 3/2 <f(x)<3\lor f(x)>3