If \(f(x) = 2x^{2} - 3x - 1\), find the value of x for which f(x) is minimum.
Answer Details
To find the value of x for which f(x) is minimum, we need to find the critical points of the function f(x) and determine whether they correspond to a minimum or maximum value.
The critical points are the values of x where the derivative of f(x) is equal to zero or does not exist.
Taking the derivative of f(x) with respect to x, we get:
$$f'(x) = 4x - 3$$
Setting f'(x) equal to zero and solving for x, we get:
$$4x - 3 = 0$$
$$x = \frac{3}{4}$$
Therefore, x = 3/4 is a critical point of f(x).
To determine whether x = 3/4 corresponds to a minimum or maximum value of f(x), we can use the second derivative test. The second derivative of f(x) is:
$$f''(x) = 4$$
Since f''(x) is positive for all values of x, we can conclude that x = 3/4 corresponds to a minimum value of f(x).
Therefore, the value of x for which f(x) is minimum is x = 3/4.