Question

let ( f ) be the function defined by ( f(x) = 3x^3 - 36x + 6 ) for ( -4 < x < 4 ). which of the following statements is true?
a ( f ) is decreasing on the interval ( (0, 4) ) because ( f(x) < 0 ) on the interval ( (0, 4) ).
b ( f ) is increasing on the interval ( (0, 4) ) because ( f(x) < 0 ) on the interval ( (0, 4) ).
c ( f ) is decreasing on the interval ( (-2, 0) ) because ( f(x) < 0 ) on the interval ( (-2, 0) ).
d ( f ) is decreasing on the interval ( (-2, 2) ) because ( f(x) < 0 ) on the interval ( (-2, 2) ).

Explanation:

Step1: Find the first derivative of \( f(x) \)

The function is \( f(x) = 3x^3 - 36x + 6 \). Using the power rule, the first derivative \( f'(x) \) is:
\( f'(x) = \frac{d}{dx}(3x^3 - 36x + 6) = 9x^2 - 36 \)

Step2: Analyze the sign of \( f'(x) \) in different intervals

First, factor \( f'(x) \): \( f'(x) = 9(x^2 - 4) = 9(x - 2)(x + 2) \)

For interval \( (0, 4) \):

Pick a test point, say \( x = 3 \). Then \( f'(3) = 9(3 - 2)(3 + 2) = 9(1)(5) = 45 > 0 \). Wait, but let's check the critical points. The critical points are where \( f'(x) = 0 \), so \( 9(x - 2)(x + 2) = 0 \) gives \( x = 2 \) or \( x = -2 \).

Wait, maybe I made a mistake earlier. Let's re - evaluate \( f'(x) \) in \( (0, 4) \). For \( x\in(0, 2) \), let's take \( x = 1 \): \( f'(1)=9(1 - 2)(1 + 2)=9(-1)(3)=-27<0 \). For \( x\in(2, 4) \), take \( x = 3 \): \( f'(3)=9(3 - 2)(3 + 2)=45>0 \).

Wait, the options:

Option A: Says \( f \) is decreasing on \( (0, 4) \) because \( f'(x)<0 \) on \( (0, 4) \). But as we saw, on \( (2, 4) \), \( f'(x)>0 \). Wait, maybe I miscalculated the derivative. Wait, \( f(x)=3x^3-36x + 6 \), so \( f'(x)=9x^2-36 \). Let's set \( f'(x)=0 \), \( 9x^2=36 \), \( x^2 = 4 \), \( x=\pm2 \).

Now, let's check option D: Interval \( (-2, 2) \). For \( x\in(-2, 2) \), let's take \( x = 0 \), \( f'(0)=9(0)^2-36=-36<0 \). Take \( x = 1 \), \( f'(1)=9 - 36=-27<0 \). Take \( x=-1 \), \( f'(-1)=9 - 36=-27<0 \). So on \( (-2, 2) \), \( f'(x)=9x^2 - 36=9(x^2 - 4) \). Since \( x\in(-2, 2) \), \( x^2<4 \), so \( x^2 - 4<0 \), and \( 9>0 \), so \( f'(x)<0 \) on \( (-2, 2) \). And when \( f'(x)<0 \), the function is decreasing.

Let's check other options:

Option A: On \( (0, 4) \), when \( x\in(0, 2) \), \( f'(x)<0 \), when \( x\in(2, 4) \), \( f'(x)>0 \). So \( f \) is not decreasing on the entire \( (0, 4) \).

Option B: If \( f'(x)<0 \), the function is decreasing, not increasing. So B is wrong.

Option C: \( f''(x) \) is the second derivative. \( f'(x)=9x^2 - 36 \), so \( f''(x)=18x \). The second derivative tells us about concavity, not about increasing or decreasing (which is determined by the first derivative). So C is wrong.

Option D: On \( (-2, 2) \), \( f'(x)=9x^2 - 36 \). Since \( x^2<4 \) for \( x\in(-2, 2) \), \( 9x^2-36=9(x^2 - 4)<0 \). And when \( f'(x)<0 \), the function is decreasing. So D is correct.

Answer:

D. \( f \) is decreasing on the interval \( (-2, 2) \) because \( f'(x) < 0 \) on the interval \( (-2, 2) \)