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We have \[ \sqrt {4x^2 - 8x + 64} \leq \abs *{x + 8} \iff 0 \leq 4x^2 - 8x + 64 \leq (x + 8)^2. \]
The left inequality can be simplified as follows: \begin {align*} 4x^2 - 8x + 64 & \geq 0 \\ x^2 - 2x + 16 & \geq 0 \\ (X - 1)^2 + 15 & \geq 0, \end {align*}
which is always true.
The right inequality can be simplified as follows: \begin {align*} 4x^2 - 8x + 64 & \leq (x + 8)^2 \\ 4x^2 - 8x + 64 & \leq x^2 + 16x + 64 \\ 3x^2 - 24 x & \leq 0 \\ x (x - 8) & \leq 0, \end {align*}
which solves to \(0 \leq x \leq 8\).
Hence, the solution to the original inequality is \(x \in [0, 8]\).
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WE have \[ \sqrt {4x^2 - 8x + 64} \leq \abs *{3x - 8} \iff 0 \leq 4x^2 - 8x + 64 \leq (3x - 8)^2. \]
The left inequality is always true from the previous part.
The right inequality can be simplified as follows: \begin {align*} 4x^2 - 8x + 64 & \leq (3x - 8)^2 \\ 4x^2 - 8x + 64 & \leq 9x^2 - 48x + 64 \\ 5x^2 - 40x & \geq 0 \\ x (x - 8) & \geq 0, \end {align*}
which solves to \(x \leq 0\) or \(x \geq 8\).
Hence, the solution to the original inequality is \(x \in (-\infty , 0] \cup [8, \infty )\).
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We have \begin {align*} \left (\sqrt {4x^2 - 8x + 64} + 2(x - 1)\right )f(x) & = \left (\sqrt {4x^2 - 8x + 64}\right )^2 - [2 (x - 1)]^2 \\ & = \left (4x^2 - 8x + 64\right ) - 4 \left (x^2 - 2x + 1\right ) \\ & = \left (4x^2 - 8x + 64\right ) - \left (4x^2 - 8x + 4\right ) \\ & = 60. \end {align*}
Hence, \[ f(x) = \frac {60}{\sqrt {4x^2 - 8x + 64} + 2 (x - 1)}. \]
As \(x \to \infty \), \(\sqrt {4x^2 - 8x + 64} \to \infty \), \(2 (x - 1) \to \infty \).
Hence, \(f(x) \to 0\) as \(x \to \infty \).
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Let \(f_1(x) = \sqrt {4x^2 - 8x + 64}\), \(f_2(x) = 2 (x - 1)\).
\(f_1(0) = \sqrt {64} = 8\), and \(f_2(0) = 2 \cdot (-1) = -2\).
We have \(f(x) = f_1(x) - f_2(x) > 0\) from the previous part, and that \(f_1(x)\) is defined for all \(x\) and is always positive.
Furthermore, \[ f_1(x) = 2 \sqrt {x^2 - 2x + 16} = 2 \sqrt {(x - 1)^2 + 15}, \] and hence \(f_1\) decreases on \((-\infty , 1)\) and increases on \((1, \infty )\), taking a minimum of \(f_1(1) = 2 \sqrt {15}\).
In terms of symmetry, we have \(f_1(1 - x) = f_1(1 + x)\) and \(f_2(1 - x) = - f_2(1 + x)\). \(f_2\) is an asymptote to \(f_1\) as \(x \to \infty \), and \(-f_2\) is an asymptote to \(f_1\) as \(x \to -\infty \).
Hence, the sketch looks as follows.
Let \(x = 3\), and we must have \(\sqrt {4 \cdot 9 - 5 \cdot 3 + 4} = \abs *{3m + c}\), and hence \(5 = \abs *{3m + c}\).
This is only achievable for \(m = \pm 2\) due to the diagram – the solution set can only be ’one-sided’ if on the other side the absolute value is eventually ’parallel’ to the curve.
We let \(m = 2\), and hence \(5 = \abs *{6 + c}\), which gives \(c = -1\) or \(c = -11\).
We would like to show that the desired value is \(c = -1\), and that \(c = -11\) does not work. \[ \sqrt {4x^2 - 5x + 4} \leq \abs *{2x - 1} \iff 0 \leq 4x^2 - 5x + 4 \leq (2x - 1)^2. \]
The left inequality can be simplified as \[ 0 \leq 4x^2 - 5x + 4 = \left (2x - \frac {5}{4}\right )^2 + \frac {39}{16}, \] and hence is always true.
The right inequality can be simplified as \begin {align*} 4x^2 - 5x + 4 & \leq (2x - 1)^2 \\ 4x^2 - 5x + 4 & \leq 4x^2 - 4x + 1 \\ x & \geq 3, \end {align*}
and hence the solution set to the whole inequality is \(x \geq 3\) as desired.
On the other hand, for the case of \(c = -11\), we have \[ \sqrt {4x^2 - 5x + 4} \leq \abs *{2x - 11} \iff 0 \leq 4x^2 - 5x + 4 \leq (2x - 11)^2, \] and the left inequality is always true by previously. However, the right inequality simplifies as \begin {align*} 4x^2 - 5x + 4 & \leq (2x - 11)^2 \\ 4x^2 - 5x + 4 & \leq 4x^2 - 44x + 121 \\ 39x & \leq 117 \\ x & \leq 3, \end {align*}
and the inequality is in the wrong direction.
Hence, a possible value of \(m\) is \(2\), and the corresponding value of \(c\) is \(-1\).
The diagram as follows shows the only possibility of the configuration.
Hence, we must have \(x^2 + px + q = mx + c\) for \(x = -5\) and \(x = 7\), and \(x^2 + px + q = -mx - c\) for \(x = 1\) and \(x = 5\). \[ \left \{ \begin {aligned} 25 - 5p + q & = -5m + c, \\ 49 + 7p + q & = 7m + c, \\ 1 + p + q & = - (m + c), \\ 25 + 5p + q & = - (5m + c). \end {aligned} \right . \]
Subtracting the first equation from the final equation gives \(10 p = -2c\), and hence \(c = -5p\).
Subtracting the first equation from the second equation gives us \(24 + 12 p = 12 m\), and hence \(m = 2 + p\).
Putting these into the third equation gives \begin {align*} q & = -m - c - ; - 1 \\ & = - (2 + p) - (-5p) - p - 1 \\ & = 3p - 3. \end {align*}
Putting all these into the final equation gives \begin {align*} 25 + 5p + (3p - 3) & = - \left [5 (2 + p) + (-5p)\right ] \\ 25 + 8p - 3 & = - (10 + 5p - 5p) \\ 22 + 8p & = -10 \\ 8p & = -32 \\ p & = -4, \end {align*}
and so \(q = -15, m = -2, c = 20\). Hence, \[ (p, q, m, c) = (-4, -15, -2, 20). \]