What is the range of f(x)=(x^3+x^2+x+1)/(|x+1|) ?

The range of f(x)=(x^3+x^2+x+1)/(|x+1|) is f(x) = 1

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range of f(x)=(x^3+x^2+x+1)/(|x+1|)

range

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

f (x) < - 2 or f (x) \geq 1

Interval Notation

(- \infty, - 2) \cup [1, \infty)

Solution steps

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

Domain of

\frac{x ^{3} + x ^{2} + x + 1}{∣ x + 1 ∣} : x < - 1 or x > - 1

Extreme Points of

\frac{x ^{3} + x ^{2} + x + 1}{∣ x + 1 ∣} :

Minimum

(0, 1)

Absolute value functions are piecewise functions

Find the function intervals:

x < - 1, x > - 1

Evaluate the function behavior for each interval

Find the range for the interval

- \infty < x < - 1 : - \infty < f (x) < - 2

Find the range for the interval

- 1 < x < \infty : 1 \leq f (x) < \infty

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

f (x) < - 2 or f (x) \geq 1

f (x) < - 2 or f (x) \geq 1

f (x) = \frac{x}{1 - x}

x^{2} - 7, x \geq 0

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

9 + x^{2}

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

What is the range of f(x)=(x^3+x^2+x+1)/(|x+1|) ?
The range of f(x)=(x^3+x^2+x+1)/(|x+1|) is f(x)<-2\lor f(x)>= 1