1. Problem: Given functions $f(x) = 4x - 1$, $g(x) = x^2 - 5$, and $h(x) = 3^x$, find: 1.i. $g(x - 2)$ in simplest form. 1.ii. $f^{-1}(11)$ (the inverse of $f$ evaluated at 11). 1.iii. $hh(1)$ (composition of $h$ with itself at 1). --- 2. Formula and rules: - To find $g(x - 2)$, substitute $x - 2$ into $g(x)$. - To find $f^{-1}(y)$, solve $y = f(x)$ for $x$. - To find $hh(1)$, compute $h(h(1))$. --- 3. Step-by-step solutions: 1.i. $$g(x - 2) = (x - 2)^2 - 5$$ Expand the square: $$= (x^2 - 4x + 4) - 5$$ Simplify: $$= x^2 - 4x - 1$$ 1.ii. Given $f(x) = 4x - 1$, find $f^{-1}(11)$. Set $y = 4x - 1$ and solve for $x$: $$y = 4x - 1$$ Add 1 to both sides: $$y + 1 = 4x$$ Divide both sides by 4: $$x = \frac{y + 1}{4}$$ Using \cancel to show division: $$x = \frac{\cancel{y + 1}}{\cancel{4}}$$ Substitute $y = 11$: $$f^{-1}(11) = \frac{11 + 1}{4} = \frac{12}{4} = 3$$ 1.iii. Calculate $hh(1) = h(h(1))$. First find $h(1)$: $$h(1) = 3^1 = 3$$ Then find $h(3)$: $$h(3) = 3^3 = 27$$ Therefore, $$hh(1) = 27$$ --- Final answers: 1.i. $g(x - 2) = x^2 - 4x - 1$ 1.ii. $f^{-1}(11) = 3$ 1.iii. $hh(1) = 27$