If x is real, then find the minimum and maximum value of the following rational function `f(x)=(x^2-3x+4)/(x^2+3x+4)`

Options

A. `1//7,7`
B. `-1//7,7`
C. `-1//7,-7`
D. `-1,1//7`

Answer

To find the minimum and maximum values of the rational function \( f(x) = \frac{x^2 - 3x + 4}{x^2 + 3x + 4} \), we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)
We start by applying the quotient rule for differentiation. The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

In our case, let \( g(x) = x^2 - 3x + 4 \) and \( h(x) = x^2 + 3x + 4 \).

Calculating the derivatives:
- \( g'(x) = 2x - 3 \)
- \( h'(x) = 2x + 3 \)

Now applying the quotient rule:

\[ f'(x) = \frac{(2x - 3)(x^2 + 3x + 4) - (x^2 - 3x + 4)(2x + 3)}{(x^2 + 3x + 4)^2} \]

### Step 2: Set the derivative equal to zero
To find critical points, we need to set the numerator of \( f'(x) \) equal to zero:

\[ (2x - 3)(x^2 + 3x + 4) - (x^2 - 3x + 4)(2x + 3) = 0 \]

### Step 3: Simplify the equation
Expanding both products and simplifying:

1. Expand \( (2x - 3)(x^2 + 3x + 4) \):
\[ 2x^3 + 6x^2 + 8x - 3x^2 - 9x - 12 = 2x^3 + 3x^2 - x - 12 \]

2. Expand \( (x^2 - 3x + 4)(2x + 3) \):
\[ 2x^3 + 3x^2 - 6x - 9 + 8 = 2x^3 + 3x^2 - 6x + 12 \]

Now, set the equation:
\[ (2x^3 + 3x^2 - x - 12) - (2x^3 + 3x^2 - 6x + 12) = 0 \]

This simplifies to:
\[ 5x - 24 = 0 \implies x = \frac{24}{5} \]

### Step 4: Evaluate \( f(x) \) at critical points and endpoints
Now we will evaluate \( f(x) \) at \( x = 2 \) and \( x = -2 \) (the critical points found from the derivative):

1. Calculate \( f(2) \):
\[ f(2) = \frac{2^2 - 3(2) + 4}{2^2 + 3(2) + 4} = \frac{4 - 6 + 4}{4 + 6 + 4} = \frac{2}{14} = \frac{1}{7} \]

2. Calculate \( f(-2) \):
\[ f(-2) = \frac{(-2)^2 - 3(-2) + 4}{(-2)^2 + 3(-2) + 4} = \frac{4 + 6 + 4}{4 - 6 + 4} = \frac{14}{2} = 7 \]

### Step 5: Determine minimum and maximum values
From our evaluations:
- The minimum value of \( f(x) \) is \( \frac{1}{7} \) at \( x = 2 \).
- The maximum value of \( f(x) \) is \( 7 \) at \( x = -2 \).

### Final Answer
- Minimum value: \( \frac{1}{7} \)
- Maximum value: \( 7 \)

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