Find the maximum and minimum value of the polynomial f(x)`=-x^(2)+x+2`.

Explanation

To find the maximum and minimum value of the polynomial \( f(x) = -x^2 + x + 2 \), we will follow these steps:

### Step 1: Find the first derivative of the polynomial
The first step is to differentiate the polynomial \( f(x) \).

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

Using the power rule, we get:

\[ f'(x) = -2x + 1 \]

### Step 2: Set the first derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ -2x + 1 = 0 \]

Solving for \( x \):

\[ 2x = 1 \implies x = \frac{1}{2} \]

### Step 3: Find the second derivative of the polynomial
Next, we find the second derivative to determine the nature of the critical point:

\[ f''(x) = \frac{d}{dx}(-2x + 1) = -2 \]

### Step 4: Analyze the second derivative
The second derivative \( f''(x) = -2 \) is less than zero. This indicates that the function is concave down at \( x = \frac{1}{2} \), which means that the critical point is a maximum.

### Step 5: Find the maximum value
Now, we substitute \( x = \frac{1}{2} \) back into the original polynomial to find the maximum value:

\[ f\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) + 2 \]

Calculating this:

\[ = -\frac{1}{4} + \frac{1}{2} + 2 \]
\[ = -\frac{1}{4} + \frac{2}{4} + \frac{8}{4} \]
\[ = \frac{-1 + 2 + 8}{4} = \frac{9}{4} \]

### Conclusion
The maximum value of the polynomial \( f(x) = -x^2 + x + 2 \) is \( \frac{9}{4} \) at \( x = \frac{1}{2} \). There is no minimum value since the polynomial opens downwards.