Find the maximum or minimum values of the following expressions on R
`2x + 5 – 3x^(2)`

Answer

To find the maximum or minimum values of the expression \( f(x) = 2x + 5 - 3x^2 \), we can follow these steps:

### Step 1: Identify the function
We start by defining the function:
\[ f(x) = 2x + 5 - 3x^2 \]

### Step 2: Find the first derivative
To find the critical points, we need to calculate the first derivative of the function:
\[ f'(x) = \frac{d}{dx}(2x + 5 - 3x^2) \]
Calculating this gives:
\[ f'(x) = 2 - 6x \]

### Step 3: Set the first derivative to zero
Next, we set the first derivative equal to zero to find the critical points:
\[ 2 - 6x = 0 \]
Solving for \( x \):
\[ 6x = 2 \quad \Rightarrow \quad x = \frac{2}{6} = \frac{1}{3} \]

### Step 4: Find the second derivative
To determine whether this critical point is a maximum or minimum, we calculate the second derivative:
\[ f''(x) = \frac{d}{dx}(2 - 6x) = -6 \]
Since \( f''(x) = -6 \) is negative, this indicates that the function has a maximum at the critical point.

### Step 5: Calculate the maximum value
Now we substitute \( x = \frac{1}{3} \) back into the original function to find the maximum value:
\[ f\left(\frac{1}{3}\right) = 2\left(\frac{1}{3}\right) + 5 - 3\left(\frac{1}{3}\right)^2 \]
Calculating this step-by-step:
\[ = \frac{2}{3} + 5 - 3\left(\frac{1}{9}\right) \]
\[ = \frac{2}{3} + 5 - \frac{1}{3} \]
Combining the fractions:
\[ = \left(\frac{2}{3} - \frac{1}{3}\right) + 5 = \frac{1}{3} + 5 = \frac{1}{3} + \frac{15}{3} = \frac{16}{3} \]

### Conclusion
The maximum value of the function \( f(x) = 2x + 5 - 3x^2 \) is:
\[ \frac{16}{3} \]
at \( x = \frac{1}{3} \).

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