GATE CS 1995 Q25a — Maxima and Minima, Mean Value Theorem, Integration

We are given the quadratic function:
$$f(x) = 2x^2 - 4x + 3$$

We can find its minimum value using two simple methods:

Method 1: Completing the Square

Rewrite the function by completing the square:
$$f(x) = 2(x^2 - 2x) + 3$$
$$f(x) = 2(x^2 - 2x + 1) - 2 + 3$$
$$f(x) = 2(x - 1)^2 + 1$$

Since $(x - 1)^2 \ge 0$ for all real $x$, the minimum value of $f(x)$ occurs when $x - 1 = 0$ (i.e., $x = 1$).
$$\text{Minimum value} = 2(0) + 1 = 1$$

Method 2: Using Calculus

Find the first derivative of $f(x)$ with respect to $x$ and set it to zero to find the critical point:
$$f'(x) = \frac{d}{dx}(2x^2 - 4x + 3) = 4x - 4$$

Setting $f'(x) = 0$ gives:
$$4x - 4 = 0 \implies x = 1$$

To confirm this is a minimum, we check the second derivative:
$$f''(x) = 4 > 0$$

Since the second derivative is positive, $x = 1$ is a point of minimum. Substituting $x = 1$ back into $f(x)$:
$$f(1) = 2(1)^2 - 4(1) + 3 = 2 - 4 + 3 = 1$$

Thus, the minimum value of the expression is 1.