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, we need to find the least value of the fraction where the denominator exceeds the square of the numerator by 16. Let's break this down step by 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
The fraction can be expressed as:
\[ f(x) = \frac{x}{x^2 + 16} \]

### Step 3: Find the Derivative
To find the minimum value of the function, we need to differentiate it. We will use the quotient rule for differentiation, which states that if we have a function \( \frac{u}{v} \), then its derivative is given by:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
Here, \( u = x \) and \( v = x^2 + 16 \).

Calculating the derivatives:
- \( u' = 1 \)
- \( v' = 2x \)

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

### Step 4: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:
\[ 16 - x^2 = 0 \]
This implies:
\[ x^2 = 16 \quad \Rightarrow \quad x = 4 \quad \text{or} \quad x = -4 \]

### Step 5: Determine the Nature of Critical Points
Next, we need to determine whether these critical points correspond to a minimum or maximum. We do this by finding the second derivative or evaluating the first derivative around these points.

### Step 6: Evaluate the Second Derivative
We can differentiate \( f'(x) \) again to find \( f''(x) \). However, for simplicity, we can evaluate \( f'(x) \) around the critical points:
- For \( x = 4 \):
- \( f'(4) = \frac{16 - 16}{(16 + 16)^2} = 0 \) (Check the sign of \( f' \) around this point)
- For \( x = -4 \):
- \( f'(-4) = \frac{16 - 16}{(16 + 16)^2} = 0 \) (Check the sign of \( f' \) around this point)

### Step 7: Evaluate the Function at Critical Points
Now we evaluate the original function \( f(x) \) at the critical points:
- For \( x = 4 \):
\[ f(4) = \frac{4}{4^2 + 16} = \frac{4}{16 + 16} = \frac{4}{32} = \frac{1}{8} \]
- For \( x = -4 \):
\[ f(-4) = \frac{-4}{(-4)^2 + 16} = \frac{-4}{16 + 16} = \frac{-4}{32} = -\frac{1}{8} \]

### Conclusion
The least value of the fraction occurs at \( x = -4 \), which gives us:
\[ \text{Least value of the fraction} = -\frac{1}{8} \]