The denominator of a fraction exceeds the square of the numberator by 16, then the least value of the fraction is

Options

A. `-(1)/(4)`
B. `-(1)/(8)`
C. `(1)/(12)`
D. `(1)/(16)`

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and provide a detailed explanation for each step.

### Step 1: Define the Variables
Let the numerator of the fraction be \( x \). According to the problem, the denominator exceeds the square of the numerator by 16. Therefore, we can express the denominator as:
\[ \text{Denominator} = x^2 + 16 \]

### Step 2: Write the Fraction
Now, we can write the fraction \( y \) as:
\[ y = \frac{x}{x^2 + 16} \]

### Step 3: Find the Derivative
To find the minimum value of the fraction, we need to differentiate \( y \) with respect to \( x \). We will use the quotient rule for differentiation, which states that if \( y = \frac{u}{v} \), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
In our case, \( u = x \) and \( v = x^2 + 16 \).

Calculating the derivatives:
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = 2x \)

Now applying the quotient rule:
\[ \frac{dy}{dx} = \frac{(x^2 + 16)(1) - (x)(2x)}{(x^2 + 16)^2} \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{x^2 + 16 - 2x^2}{(x^2 + 16)^2} = \frac{16 - x^2}{(x^2 + 16)^2} \]

### Step 4: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:
\[ 16 - x^2 = 0 \]
Solving for \( x \):
\[ x^2 = 16 \implies x = \pm 4 \]

### Step 5: Evaluate the Fraction at Critical Points
Now we will evaluate the fraction \( y \) at \( x = 4 \) and \( x = -4 \).

1. For \( x = 4 \):
\[ y = \frac{4}{4^2 + 16} = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8} \]

2. For \( x = -4 \):
\[ y = \frac{-4}{(-4)^2 + 16} = \frac{-4}{16 + 16} = \frac{-4}{32} = -\frac{1}{8} \]

### Step 6: Determine the Minimum Value
From the evaluations:
- At \( x = 4 \), \( y = \frac{1}{8} \)
- At \( x = -4 \), \( y = -\frac{1}{8} \)

The least value of the fraction is:
\[ \text{Minimum value of } y = -\frac{1}{8} \]

### Conclusion
Thus, the least value of the fraction is:
\[ \boxed{-\frac{1}{8}} \]