Find the absolute maximum and minimum, if either exists, for \( f(x)=x^{2}-4 x+2 \). Find the first derivative of \( f \). \( f^{\prime}(x)=\square \)

To find the first derivative of the function \( f(x) = x^2 - 4x + 2 \), we apply basic differentiation rules. The derivative of \( x^2 \) is \( 2x \), the derivative of \( -4x \) is \( -4 \), and the derivative of the constant \( 2 \) is \( 0 \). Therefore, \[ f^{\prime}(x) = 2x - 4. \] Now, to find the absolute maximum and minimum of \( f(x) \), we need to determine the critical points by setting the first derivative equal to zero: \[ 2x - 4 = 0 \implies 2x = 4 \implies x = 2. \] Next, we evaluate \( f(x) \) at \( x = 2 \): \[ f(2) = 2^2 - 4(2) + 2 = 4 - 8 + 2 = -2. \] Since this is a quadratic function that opens upwards (the coefficient of \( x^2 \) is positive), we can conclude that the parabola has a minimum at \( x = 2 \), with no absolute maximum. Therefore, the absolute minimum value is \( -2 \) at \( x = 2 \), and there is no absolute maximum. So, to fill in the box: \( f^{\prime}(x) = 2x - 4. \)