What is the range of ((x^3-2x^2-3x))/(x-3) ?

The range of ((x^3-2x^2-3x))/(x-3) is f(x)>=-1/4

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range of ((x^3-2x^2-3x))/(x-3)

range

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

f (x) \geq - \frac{1}{4}

Interval Notation

[- \frac{1}{4}, \infty)

Solution steps

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

Domain of

\frac{x ^{3} - 2 x ^{2} - 3 x}{x - 3} : x < 3 or x > 3

Simplify

\frac{x ^{3} - 2 x ^{2} - 3 x}{x - 3} : x (x + 1)

Extreme Points of

x (x + 1) :

Minimum

(- \frac{1}{2}, - \frac{1}{4})

Find the range for the interval

- \infty < x < 3 : - \frac{1}{4} \leq f (x) < \infty

Find the range for the interval

3 < x < \infty : 12 < f (x) < \infty

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

f (x) \geq - \frac{1}{4} or f (x) > 12

f (x) \geq - \frac{1}{4}

f (x) = (\frac{x - 2}{5})^{\frac{1}{5}}

f (x) = \frac{( x - 5 )}{( x - 7 )}

2 x - y = - 4

sin^{2} (θ)

f (x) = \frac{x - 5}{x + 5}

What is the range of ((x^3-2x^2-3x))/(x-3) ?
The range of ((x^3-2x^2-3x))/(x-3) is f(x)>=-1/4