Finding supremum and infimum of family of finite real valued functions

Question

Let $\mathcal{F}$ be a family of finite real valued functions $f:[-3,2]\to \mathbb{R}$. Suppose I have $\mathcal{F}=\{x^2, x+1, sin\frac{\pi}{2}x\}$ and $$g(x)=\sup_{f\in\mathcal{F}} f(x),\quad h(x)=\inf_{f\in\mathcal{F}} f(x).$$ Is it correct that $$g(x)=\sup_{f\in\mathcal{F}} f(x)=x^2, \quad h(x)=\inf_{f\in\mathcal{F}} f(x)=x +1$$ for $x \in [-3,2]$? I get the $g(x)$ and $h(x)$ from this plots:

Sorry if this question maybe too easy, but I am a bit confused because it's first time I am searching for a supremum and infimum of family of finite real valued functions. Thanks for any help.

Answer 1

$x^2$ and $x+1$ coincide in $x_1=\frac{1-\sqrt{5}}{2}$ and $x_2=\frac{1+\sqrt{5}}{2}.$ Hence

$g(x)=x^2$ in $[-3, x_1].$

$g(x)=x+1$ in $[x_1, x_2].$

$g(x)=x^2$ in $[x_2, 2].$

Answer 2

Let $x_3$ be such that $x_3+1=\sin(x_3\pi/2).$

According to the graph, we have

$h(x)=x+1$ in $[-3, x_3].$

$h(x)=\sin(x\pi/2)$ in $[x_3, 0].$

$h(x)=x^2$ in $[0, 1].$

$h(x)=\sin(x\pi/2)$ in $[1, 2].$