At what values of `x in R` the following quadratic expressions have maximum or minimum values
i) `x^(2)+5x+6` ii) `2x-x^(2)+7`

Answer

To solve the problem of finding the values of \( x \) at which the quadratic expressions have maximum or minimum values, we will go through each expression step by step.

### i) For the expression \( f(x) = x^2 + 5x + 6 \)

**Step 1: Find the first derivative \( f'(x) \)**
To find the critical points where the function may have maximum or minimum values, we first compute the first derivative of the function:
\[ f'(x) = \frac{d}{dx}(x^2 + 5x + 6) = 2x + 5 \]

**Step 2: Set the first derivative to zero**
Next, we set the first derivative equal to zero to find the critical points:
\[ 2x + 5 = 0 \]
Solving for \( x \):
\[ 2x = -5 \quad \Rightarrow \quad x = -\frac{5}{2} \]

**Step 3: Find the second derivative \( f''(x) \)**
To determine whether this critical point is a maximum or minimum, we compute the second derivative:
\[ f''(x) = \frac{d^2}{dx^2}(x^2 + 5x + 6) = 2 \]

**Step 4: Analyze the second derivative**
Since \( f''(x) = 2 \) is positive, this indicates that the function \( f(x) \) has a minimum value at \( x = -\frac{5}{2} \).

### ii) For the expression \( g(x) = 2x - x^2 + 7 \)

**Step 1: Find the first derivative \( g'(x) \)**
We compute the first derivative of this function:
\[ g'(x) = \frac{d}{dx}(2x - x^2 + 7) = 2 - 2x \]

**Step 2: Set the first derivative to zero**
Next, we set the first derivative equal to zero to find the critical points:
\[ 2 - 2x = 0 \]
Solving for \( x \):
\[ 2 = 2x \quad \Rightarrow \quad x = 1 \]

**Step 3: Find the second derivative \( g''(x) \)**
To determine whether this critical point is a maximum or minimum, we compute the second derivative:
\[ g''(x) = \frac{d^2}{dx^2}(2x - x^2 + 7) = -2 \]

**Step 4: Analyze the second derivative**
Since \( g''(x) = -2 \) is negative, this indicates that the function \( g(x) \) has a maximum value at \( x = 1 \).

### Summary of Results
- For \( f(x) = x^2 + 5x + 6 \), the minimum value occurs at \( x = -\frac{5}{2} \).
- For \( g(x) = 2x - x^2 + 7 \), the maximum value occurs at \( x = 1 \).