Find the maximum or minimum values of the following expressions on R
i) `x^(2) + 6x – 27` ii) `3x^(2) + 2x + 7` iii) `x^(2) - 12x + 32` iv) `2x^(2) + 3x + 1`

Answer

To find the maximum or minimum values of the given quadratic expressions, we will use the concept of derivatives. For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex form gives us the maximum or minimum value depending on the sign of \( a \). If \( a > 0 \), the function has a minimum value; if \( a < 0 \), it has a maximum value.

### Step-by-step Solutions:

### i) For the expression \( f(x) = x^2 + 6x - 27 \)

1. **Find the derivative**:
\[ f'(x) = 2x + 6 \]

2. **Set the derivative to zero to find critical points**:
\[ 2x + 6 = 0 \implies x = -3 \]

3. **Second derivative test**:
\[ f''(x) = 2 \quad (\text{which is positive}) \]
Since \( f''(x) > 0 \), \( f(x) \) has a minimum at \( x = -3 \).

4. **Calculate the minimum value**:
\[ f(-3) = (-3)^2 + 6(-3) - 27 = 9 - 18 - 27 = -36 \]

**Minimum value of \( f(x) \) is -36.**

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### ii) For the expression \( f(x) = 3x^2 + 2x + 7 \)

1. **Find the derivative**:
\[ f'(x) = 6x + 2 \]

2. **Set the derivative to zero**:
\[ 6x + 2 = 0 \implies x = -\frac{1}{3} \]

3. **Second derivative test**:
\[ f''(x) = 6 \quad (\text{which is positive}) \]
Since \( f''(x) > 0 \), \( f(x) \) has a minimum at \( x = -\frac{1}{3} \).

4. **Calculate the minimum value**:
\[ f\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right)^2 + 2\left(-\frac{1}{3}\right) + 7 = 3\left(\frac{1}{9}\right) - \frac{2}{3} + 7 \]
\[ = \frac{1}{3} - \frac{2}{3} + 7 = -\frac{1}{3} + 7 = \frac{20}{3} \]

**Minimum value of \( f(x) \) is \( \frac{20}{3} \).**

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### iii) For the expression \( f(x) = x^2 - 12x + 32 \)

1. **Find the derivative**:
\[ f'(x) = 2x - 12 \]

2. **Set the derivative to zero**:
\[ 2x - 12 = 0 \implies x = 6 \]

3. **Second derivative test**:
\[ f''(x) = 2 \quad (\text{which is positive}) \]
Since \( f''(x) > 0 \), \( f(x) \) has a minimum at \( x = 6 \).

4. **Calculate the minimum value**:
\[ f(6) = 6^2 - 12 \cdot 6 + 32 = 36 - 72 + 32 = -4 \]

**Minimum value of \( f(x) \) is -4.**

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### iv) For the expression \( f(x) = 2x^2 + 3x + 1 \)

1. **Find the derivative**:
\[ f'(x) = 4x + 3 \]

2. **Set the derivative to zero**:
\[ 4x + 3 = 0 \implies x = -\frac{3}{4} \]

3. **Second derivative test**:
\[ f''(x) = 4 \quad (\text{which is positive}) \]
Since \( f''(x) > 0 \), \( f(x) \) has a minimum at \( x = -\frac{3}{4} \).

4. **Calculate the minimum value**:
\[ f\left(-\frac{3}{4}\right) = 2\left(-\frac{3}{4}\right)^2 + 3\left(-\frac{3}{4}\right) + 1 \]
\[ = 2\left(\frac{9}{16}\right) - \frac{9}{4} + 1 = \frac{18}{16} - \frac{36}{16} + \frac{16}{16} = \frac{18 - 36 + 16}{16} = \frac{-2}{16} = -\frac{1}{8} \]

**Minimum value of \( f(x) \) is \( -\frac{1}{8} \).**

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### Summary of Results:
1. Minimum of \( x^2 + 6x - 27 \) is -36.
2. Minimum of \( 3x^2 + 2x + 7 \) is \( \frac{20}{3} \).
3. Minimum of \( x^2 - 12x + 32 \) is -4.
4. Minimum of \( 2x^2 + 3x + 1 \) is \( -\frac{1}{8} \).