Question Type 2: Determining if a specific relation is a function Bootcamps

Given the one-sided relation $y = \sqrt{1 - x^2},$ determine if this defines $y$ as a function on $[-1,1]$ and justify your answer with reference to the definition of a function.

Restrict the relation $x^2+y^2=1$ to the upper semicircle and write $y$ explicitly as a function of $x$. State the domain of this function.

Restrict the relation $x^2+y^2=1$ to the lower semicircle and write $y$ explicitly as a function of $x$. State the domain of this function.

Determine whether the implicit relation $x^2+y^2=1$ defines $y$ as a function of $x$ over the interval $[-1,1]$. Use the vertical line test to justify your answer.

Consider the mapping that assigns to each $x$ in $[-1,1]$ the two values $y = \pm\sqrt{1 - x^2}$. Does this mapping define a function? Explain.

Parametrize the circle by $x=\cos t,\quad y=\sin t$ for $t\in[0,\pi/2]$. Does this parametrization define $y$ as a function of $x$ on that interval? Justify using the relation between $x$ and $y$.

Solve the equation $x^2+y^2=1$ for $y$ and discuss whether the solutions define $y$ as a single-valued function on the domain $[-1,1]$.

Define $f(x)=\sqrt{1 - x^2}$ and $g(x)=-\sqrt{1 - x^2}$ on $[-1,1]$. Are $f$ and $g$ each functions on $[-1,1]$? Discuss whether either has an inverse that satisfies the circle equation.

Now consider the same parametrization for $t\in[0,\pi]$. Does this define $y$ as a single-valued function of $x$ on $[-1,1]$? Explain.

Explain why there is no single real-valued continuous function $y=f(x)$ defined on $[-1,1]$ whose graph is the entire circle $x^2+y^2=1$.

Use the implicit function theorem on $F(x,y)=x^2+y^2-1=0$ to determine where $y$ can be expressed as a differentiable function of $x$. Specify the regions in the plane.