(a) Two functions, f and g, are defined by \(f : x \to 2x^{2} - 1\) and \(g : x \to 3x + 2\) where x is a real number. (i) If \(f(x - 1) - 7 = 0\), find the...

Answer Details

(a)(i) \(f(x) = 2x^2 - 1\), so \(f(x-1) = 2(x-1)^2 - 1\).

\[2(x-1)^2 - 1 - 7 = 0 \;\Rightarrow\; 2(x-1)^2 = 8 \;\Rightarrow\; (x-1)^2 = 4\]

\[x - 1 = \pm 2 \;\Rightarrow\; x = 3 \text{ or } x = -1\]

(ii) Evaluate each term:

• \(f\!\left(-\tfrac12\right) = 2\left(\tfrac14\right) - 1 = -\tfrac12\)
• \(g(3) = 3(3) + 2 = 11\)
• \(f(4) = 2(16) - 1 = 31\)
• \(g(5) = 3(5) + 2 = 17\)

\[\frac{f\!\left(-\tfrac12\right)\cdot g(3)}{f(4) - g(5)} = \frac{\left(-\tfrac12\right)(11)}{31 - 17} = \frac{-\tfrac{11}{2}}{14} = -\frac{11}{28}\]

(b) \(m \ast n = \dfrac{-n}{m^2 + 1}\). Test with \(m = -3,\, n = -10\):

\[(-3)\ast(-10) = \frac{-(-10)}{(-3)^2 + 1} = \frac{10}{10} = 1\]

\[(-10)\ast(-3) = \frac{-(-3)}{(-10)^2 + 1} = \frac{3}{101}\]

Since \(1 \neq \dfrac{3}{101}\), we have \((-3)\ast(-10) \neq (-10)\ast(-3)\). Therefore the operation \(\ast\) is not commutative.