Question

let ( f ) be the function defined by ( f(x) = \frac{x^3}{3} - \frac{x^2}{2} - 6x ). on which open intervals is ( f ) decreasing?
a ( -2 < x < 3 ) only
b ( x < -2 ) and ( x > 3 )
c ( x < \frac{1}{2} ) only
d there are no such intervals.

Explanation:

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

To determine where a function is decreasing, we first find its derivative. Given \( f(x)=\frac{x^{3}}{3}-\frac{x^{2}}{2}-6x \), we use the power rule for differentiation. The power rule states that if \( y = x^{n} \), then \( y'=nx^{n - 1} \).

For the first term \( \frac{x^{3}}{3} \), the derivative is \( \frac{3x^{2}}{3}=x^{2} \).

For the second term \( -\frac{x^{2}}{2} \), the derivative is \( -\frac{2x}{2}=-x \).

For the third term \( - 6x \), the derivative is \( -6 \).

So, the derivative \( f'(x)=x^{2}-x - 6 \).

Step2: Factor the derivative

We factor the quadratic expression \( x^{2}-x - 6 \). We need two numbers that multiply to \( - 6 \) and add up to \( -1 \). The numbers are \( -3 \) and \( 2 \).

So, \( x^{2}-x - 6=(x - 3)(x + 2) \).

Step3: Find critical points

Critical points occur where \( f'(x)=0 \) or \( f'(x) \) is undefined. Since \( f'(x) \) is a polynomial, it is defined everywhere. We set \( f'(x) = 0 \):

\( (x - 3)(x + 2)=0 \)

This gives us \( x - 3=0 \) or \( x + 2=0 \), so \( x = 3 \) or \( x=-2 \).

Step4: Test intervals

We divide the real number line into intervals based on the critical points \( x=-2 \) and \( x = 3 \). The intervals are \( (-\infty,-2) \), \( (-2,3) \), and \( (3,\infty) \).

• For the interval \( (-\infty,-2) \), we can test \( x=-3 \). Substitute into \( f'(x)=(x - 3)(x + 2) \): \( (-3 - 3)(-3 + 2)=(-6)(-1)=6>0 \). So, \( f(x) \) is increasing on \( (-\infty,-2) \).
• For the interval \( (-2,3) \), we can test \( x = 0 \). Substitute into \( f'(x)=(0 - 3)(0 + 2)=(-3)(2)=-6<0 \). So, \( f(x) \) is decreasing on \( (-2,3) \).
• For the interval \( (3,\infty) \), we can test \( x = 4 \). Substitute into \( f'(x)=(4 - 3)(4 + 2)=(1)(6)=6>0 \). So, \( f(x) \) is increasing on \( (3,\infty) \).

Answer:

A. \( -2 < x < 3 \) only