Find the maximum or minimum values, if any, of the following functions without using the derivatives :
`f(x)=(2x-1)^(2)+3`

Answer

To find the maximum or minimum values of the function \( f(x) = (2x - 1)^2 + 3 \) without using derivatives, we can analyze the function step by step.

### Step 1: Rewrite the Function
We start with the function:
\[ f(x) = (2x - 1)^2 + 3 \]
This function is in the form of a quadratic equation, specifically a parabola.

### Step 2: Identify the Vertex
A quadratic function of the form \( a(x - h)^2 + k \) has its vertex at the point \( (h, k) \). In our case, we can rewrite the function as:
\[ f(x) = (2x - 1)^2 + 3 \]
Here, \( (2x - 1)^2 \) reaches its minimum value when \( 2x - 1 = 0 \).

### Step 3: Solve for \( x \)
Setting \( 2x - 1 = 0 \):
\[ 2x = 1 \implies x = \frac{1}{2} \]

### Step 4: Find the Minimum Value
Now substitute \( x = \frac{1}{2} \) back into the function to find the minimum value:
\[ f\left(\frac{1}{2}\right) = (2 \cdot \frac{1}{2} - 1)^2 + 3 = (1 - 1)^2 + 3 = 0 + 3 = 3 \]

### Step 5: Conclusion
Thus, the minimum value of the function \( f(x) \) occurs at \( x = \frac{1}{2} \) and is equal to:
\[ \text{Minimum value} = 3 \]
Since the parabola opens upwards (as indicated by the positive coefficient of the squared term), there is no maximum value for this function.

### Summary
- Minimum value of \( f(x) \) is \( 3 \) at \( x = \frac{1}{2} \).
- There is no maximum value.