Find the maximum or minimum values of the following expressions
i) `2x -7 – 5x^(2)` ii) `3x^(2)+2x+11` iii) `ax^(2)+bx+a, a,b in R, a != 0` iv) `x^(2)-x+7`

Answer

Let's solve the given expressions step by step to find their maximum or minimum values.

### i) Expression: \( f(x) = 2x - 7 - 5x^2 \)

1. **Find the first derivative**:
\[ f'(x) = \frac{d}{dx}(2x - 7 - 5x^2) = 2 - 10x \]

2. **Set the first derivative to zero**:
\[ 2 - 10x = 0 \implies 10x = 2 \implies x = \frac{1}{5} \]

3. **Find the second derivative**:
\[ f''(x) = \frac{d^2}{dx^2}(2x - 7 - 5x^2) = -10 \]
Since \( f''(x) < 0 \), this indicates a maximum.

4. **Calculate the maximum value by substituting \( x = \frac{1}{5} \) back into the original function**:
\[ f\left(\frac{1}{5}\right) = 2\left(\frac{1}{5}\right) - 7 - 5\left(\frac{1}{5}\right)^2 \]
\[ = \frac{2}{5} - 7 - 5 \cdot \frac{1}{25} \]
\[ = \frac{2}{5} - 7 - \frac{1}{5} = \frac{2 - 35 - 1}{5} = \frac{-34}{5} \]
**Maximum value**: \( -\frac{34}{5} \)

### ii) Expression: \( f(x) = 3x^2 + 2x + 11 \)

1. **Find the first derivative**:
\[ f'(x) = \frac{d}{dx}(3x^2 + 2x + 11) = 6x + 2 \]

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

3. **Find the second derivative**:
\[ f''(x) = \frac{d^2}{dx^2}(3x^2 + 2x + 11) = 6 \]
Since \( f''(x) > 0 \), this indicates a minimum.

4. **Calculate the minimum value by substituting \( x = -\frac{1}{3} \) back into the original function**:
\[ f\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right)^2 + 2\left(-\frac{1}{3}\right) + 11 \]
\[ = 3 \cdot \frac{1}{9} - \frac{2}{3} + 11 = \frac{1}{3} - \frac{2}{3} + 11 = -\frac{1}{3} + 11 = \frac{32}{3} \]
**Minimum value**: \( \frac{32}{3} \)

### iii) Expression: \( f(x) = ax^2 + bx + a \) (where \( a, b \in \mathbb{R}, a \neq 0 \))

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

2. **Set the first derivative to zero**:
\[ 2ax + b = 0 \implies x = -\frac{b}{2a} \]

3. **Find the second derivative**:
\[ f''(x) = 2a \]
- If \( a > 0 \), then \( f''(x) > 0 \) (minimum).
- If \( a < 0 \), then \( f''(x) < 0 \) (maximum).

4. **Calculate the extrema value**:
\[ f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + a \]
\[ = a\frac{b^2}{4a^2} - \frac{b^2}{2a} + a = \frac{b^2}{4a} - \frac{2b^2}{4a} + a = a - \frac{b^2}{4a} \]
- If \( a > 0 \): Minimum value is \( a - \frac{b^2}{4a} \).
- If \( a < 0 \): Maximum value is \( a - \frac{b^2}{4a} \).

### iv) Expression: \( f(x) = x^2 - x + 7 \)

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

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

3. **Find the second derivative**:
\[ f''(x) = 2 \]
Since \( f''(x) > 0 \), this indicates a minimum.

4. **Calculate the minimum value by substituting \( x = \frac{1}{2} \) back into the original function**:
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 7 \]
\[ = \frac{1}{4} - \frac{1}{2} + 7 = \frac{1}{4} - \frac{2}{4} + 7 = -\frac{1}{4} + 7 = \frac{27}{4} \]
**Minimum value**: \( \frac{27}{4} \)

### Summary of Results:
- i) Maximum value: \( -\frac{34}{5} \)
- ii) Minimum value: \( \frac{32}{3} \)
- iii) Minimum/Maximum value: \( a - \frac{b^2}{4a} \) (depends on the sign of \( a \))
- iv) Minimum value: \( \frac{27}{4} \)