The function `y=(x)/(x^2 - 6x -16 )` is a decreasing function in R-{-2,8}. Is this statement true ?

Answer

To determine whether the function \( y = \frac{x}{x^2 - 6x - 16} \) is a decreasing function in the interval \( \mathbb{R} - \{-2, 8\} \), we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we need to find its derivative \( f'(x) \). The function is in the form of \( \frac{u}{v} \), where \( u = x \) and \( v = x^2 - 6x - 16 \). We will use the quotient rule for differentiation:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Calculating \( u' \) and \( v' \):
- \( u' = 1 \)
- \( v' = 2x - 6 \)

Now substituting into the quotient rule:

\[ f'(x) = \frac{(1)(x^2 - 6x - 16) - (x)(2x - 6)}{(x^2 - 6x - 16)^2} \]

### Step 2: Simplify the derivative
Now we simplify the numerator:

\[ f'(x) = \frac{x^2 - 6x - 16 - (2x^2 - 6x)}{(x^2 - 6x - 16)^2} \]
\[ = \frac{x^2 - 6x - 16 - 2x^2 + 6x}{(x^2 - 6x - 16)^2} \]
\[ = \frac{-x^2 - 16}{(x^2 - 6x - 16)^2} \]

### Step 3: Determine where the derivative is less than or equal to zero
For the function to be decreasing, we need \( f'(x) \leq 0 \):

\[ \frac{-x^2 - 16}{(x^2 - 6x - 16)^2} \leq 0 \]

The denominator \( (x^2 - 6x - 16)^2 \) is always positive except where it is undefined (at the roots of the denominator). The numerator \( -x^2 - 16 \) is always negative since \( -x^2 \) is non-positive and subtracting 16 makes it negative. Thus, the entire fraction is less than or equal to zero for all \( x \) except where the function is undefined.

### Step 4: Identify points of discontinuity
The function is undefined at the points where the denominator is zero:

\[ x^2 - 6x - 16 = 0 \]

Factoring gives:

\[ (x - 8)(x + 2) = 0 \]

Thus, \( x = 8 \) and \( x = -2 \) are points of discontinuity.

### Step 5: Conclusion
The function \( y = \frac{x}{x^2 - 6x - 16} \) is decreasing for all \( x \) in \( \mathbb{R} \) except at the points \( -2 \) and \( 8 \). Therefore, the statement that the function is decreasing in \( \mathbb{R} - \{-2, 8\} \) is **true**.