1. Stated the problem: Given the function $m(x) = x^2 - 4x + 3$ with domain $x \geq p$, find the value of $p$ so that $m$ has an inverse function. Also, given $g^{-1}h(x) = 5x + 4$ and $fg(x) = 1 + 2x$, find $fh(5)$. 2. To have an inverse function, $m(x)$ must be one-to-one on its domain. The function $m(x) = x^2 - 4x + 3$ is a parabola opening upward with vertex at: $$x_{vertex} = \frac{-b}{2a} = \frac{4}{2} = 2$$ Since the parabola is symmetric about $x=2$, to make $m$ invertible, restrict domain to either $x \geq 2$ or $x \leq 2$. 3. The condition $x \geq p$ must cover only one branch. If $p \geq 2$, then $m$ on $[p, \infty)$ is one-to-one. 4. Therefore, the smallest $p$ for invertibility is $p = 2$. 5. Next, given $g^{-1}h(x) = 5x + 4$ and $fg(x) = 1 + 2x$, find $fh(5)$. 6. Note $g^{-1}h = 5x + 4$ means $h = g(5x + 4)$. Then: $$fh(x) = f(g(5x + 4))$$ 7. Replace $x$ by 5: $$fh(5) = f(g(5 \cdot 5 + 4)) = f(g(25 + 4)) = f(g(29))$$ 8. Since $fg(x) = 1 + 2x$, then $f(g(29)) = 1 + 2 \cdot 29 = 1 + 58 = 59$. **Final answers:** - $p = 2$ - $fh(5) = 59$