1. **State the problem:** We are given two functions: $$f(x) = 2x + 3$$ and $$g(x) = -\frac{2}{3}x + 4$$ We need to find: a) The composition $g \circ f(x)$. b) The inverse function $g^{-1}(x)$. c) The value of $m$ such that $f(m) = 2[g \circ f(1)] - \frac{1}{3}$. --- 2. **Find $g \circ f(x)$:** By definition, $g \circ f(x) = g(f(x))$. Substitute $f(x)$ into $g$: $$g(f(x)) = -\frac{2}{3}(2x + 3) + 4$$ Distribute: $$= -\frac{2}{3} \times 2x - \frac{2}{3} \times 3 + 4 = -\frac{4}{3}x - 2 + 4$$ Simplify constants: $$= -\frac{4}{3}x + 2$$ --- 3. **Find $g^{-1}(x)$:** Start with: $$y = -\frac{2}{3}x + 4$$ Swap $x$ and $y$ to find inverse: $$x = -\frac{2}{3}y + 4$$ Solve for $y$: $$x - 4 = -\frac{2}{3}y$$ Multiply both sides by $-\frac{3}{2}$: $$y = \cancel{-\frac{3}{2}}(x - 4) \times \cancel{-1} = -\frac{3}{2}x + 6$$ So, $$g^{-1}(x) = -\frac{3}{2}x + 6$$ --- 4. **Find $m$ such that $f(m) = 2[g \circ f(1)] - \frac{1}{3}$:** First, find $f(1)$: $$f(1) = 2(1) + 3 = 5$$ Next, find $g \circ f(1) = g(f(1)) = g(5)$: $$g(5) = -\frac{2}{3} \times 5 + 4 = -\frac{10}{3} + 4 = -\frac{10}{3} + \frac{12}{3} = \frac{2}{3}$$ Calculate the right side: $$2 \times \frac{2}{3} - \frac{1}{3} = \frac{4}{3} - \frac{1}{3} = 1$$ So, $$f(m) = 1$$ Recall $f(m) = 2m + 3$, set equal to 1: $$2m + 3 = 1$$ Solve for $m$: $$2m = 1 - 3 = -2$$ $$m = \frac{\cancel{2}m}{\cancel{2}} = \frac{-2}{2} = -1$$ --- **Final answers:** a) $$g \circ f(x) = -\frac{4}{3}x + 2$$ b) $$g^{-1}(x) = -\frac{3}{2}x + 6$$ c) $$m = -1$$