Function f(x) = x^3 - 3x + 3 is (A) Increasing in the interval (-1, 1) (B) Increasing in the interval (1, ∞) (C) Decreasing in the interval (-1, 1)…

To determine where $f(x) = x^3 - 3x + 3$f(x)=x3−3x+3 is increasing or decreasing, the derivative indicates the slope at any point. Positive slope means increasing, negative slope means decreasing.

$$f'(x) = 3x^2 - 3$$f′(x)=3x2−3

$$f'(x) = 3(x^2 - 1)$$f′(x)=3(x2−1)

$$f'(x) = 3(x - 1)(x + 1)$$f′(x)=3(x−1)(x+1)

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Setting $f'(x) = 0$f′(x)=0:

$$3(x - 1)(x + 1) = 0$$3(x−1)(x+1)=0

Critical points: $x = -1$x=−1 and $x = 1$x=1

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Testing the sign of $f'(x)$f′(x) in each interval:

For $(-\infty, -1)$(−∞,−1), test $x = -2$x=−2:

$$f'(-2) = 3(-2 - 1)(-2 + 1)$$f′(−2)=3(−2−1)(−2+1)

$$= 3(-3)(-1)$$=3(−3)(−1)

$$= 9 > 0$$=9>0

Function is increasing.

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For $(-1, 1)$(−1,1), test $x = 0$x=0:

$$f'(0) = 3(0 - 1)(0 + 1)$$f′(0)=3(0−1)(0+1)

$$= 3(-1)(1)$$=3(−1)(1)

$$= -3 < 0$$=−3<0

Function is decreasing.

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For $(1, \infty)$(1,∞), test $x = 2$x=2:

$$f'(2) = 3(2 - 1)(2 + 1)$$f′(2)=3(2−1)(2+1)

$$= 3(1)(3)$$=3(1)(3)

$$= 9 > 0$$=9>0

Function is increasing.

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Evaluating each option:

(A) Increasing in $(-1, 1)$(−1,1): False (the function is decreasing)

(B) Increasing in $(1, \infty)$(1,∞): True

(C) Decreasing in $(-1, 1)$(−1,1): True

(D) Increasing in $(-\infty, -1) \cup (1, \infty)$(−∞,−1)∪(1,∞): True

The correct statements are (B), (C), and (D).