The function \(f\) is defined by \(f(x) = -2x^2 + 12x - 3\) for \(x \in \mathbb{R}\).
(i) Express \(-2x^2 + 12x - 3\) in the form \(-2(x+a)^2 + b\), where \(a\) and \(b\) are constants.
(ii) State the greatest value of \(f(x)\).
The function \(g\) is defined by \(g(x) = 2x + 5\) for \(x \in \mathbb{R}\).
(iii) Find the values of \(x\) for which \(gf(x) + 1 = 0\).
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Solution
(i) To express \(-2x^2 + 12x - 3\) in the form \(-2(x+a)^2 + b\), we complete the square:
\(-2x^2 + 12x - 3 = -2(x^2 - 6x) - 3\)
Complete the square for \(x^2 - 6x\):
\(x^2 - 6x = (x-3)^2 - 9\)
Substitute back:
\(-2((x-3)^2 - 9) - 3 = -2(x-3)^2 + 18 - 3 = -2(x-3)^2 + 15\)
Thus, \(a = -3\) and \(b = 15\).
(ii) The greatest value of \(f(x)\) is the maximum value of \(-2(x-3)^2 + 15\), which occurs when \((x-3)^2 = 0\). Therefore, the greatest value is 15.
(iii) To find \(x\) for which \(gf(x) + 1 = 0\), first find \(gf(x)\):
\(gf(x) = g(f(x)) = 2(-2x^2 + 12x - 3) + 5 = -4x^2 + 24x - 6 + 5 = -4x^2 + 24x - 1\)
Set \(gf(x) + 1 = 0\):
\(-4x^2 + 24x - 1 + 1 = 0\)
\(-4x^2 + 24x = 0\)
Factor out \(-4x\):
\(-4x(x - 6) = 0\)
Thus, \(x = 0\) or \(x = 6\).
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