Grades 11-12 Video Solutions 2021-video 2021 11-12/29

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Video Summary

To find the minimum possible value of \( m(k) \), we consider the maximum of the absolute value of the function \( 4x^2 - 4x + k \) over the interval \([-1, 1]\). The parabola's maximum can occur at the endpoints or the vertex. Using the vertex formula \( x = -\frac{b}{2a} \), we find the vertex at \( x = \frac{1}{2} \). Substituting \( x = -1, 1, \frac{1}{2} \) into \( 4x^2 - 4x + k \), we determine the absolute values of expressions in terms of \( k \). By setting these equal, the minimum occurs when \( k = -\frac{7}{2} \), giving \( m(k) = \frac{9}{2} \).

Keywords

minimum value

absolute value

parabola

vertex

interval