Exercise 1.1.7 (7-14)

Solution. Solutions to these equations are all pairs $(x,y)$ which satisfy the given formula. If the set of such solutions does not fail the vertical test, i.e., there are no two pairs $(x,y_1)$ and $(x,y_2)$ with $y_1\ne y_2$, then we can equivalently represent it as $(x,f(x))$, i.e., a graph of some function $f$.

7.

To show that the graph of this equation is a graph of some function, let’s perform the Vertical Line Test by solving this equation for $y$: $$3x-5y=7;5y=3x-7;y=\frac{3x-7}{5}=0.6x-1.4.$$

We see that for all $x$ there is exactly one $y$ that satisfies the equation $y=0.6x-1.4$. In other words, the solutions of this equation define $y$ as the function of $x$.

8.

To show that the graph of this equation is a graph of some function, let’s perform the Vertical Line Test by solving this equation for $y$: $$3x^2-2y=5;2y=3x^2-5;y=\frac{3x^2-5}{2}=1.5x^2-2.5.$$

We see that for all $x$ there is exactly one $y$ that satisfies the equation $y=1.5x^2-2.5$. In other words, this equation defines $y$ as the function of $x$ because for every $x$ there is only one output assigned.

9.

This equation does not induce a function because its graph fails the Vertical Line Test, i.e., two points with the same $x$ belong to the solution set of this equation: $(1,1)$ and $(1,5)$. In other words, a vertical line through $x=1$ crosses the graph of this equation at two points.

10.

The equation $2\mathit{xy}+5y^2=4$ does not induce a function because its graph fails the Vertical Line Test. If $x=0,$ we get $5y^2=4,$ which has two solutions in $y$. Therefore the graph of this equation fails the Vertical Line Test because for one $x$ there are more than 1 $y$-coordinates.

11.

To show that the graph of this equation is a graph of some function, let’s perform the Vertical Line Test by solving this equation for $y$: $${(y+3)}^3+1=2x$$

$${(y+3)}^3=2x-1$$

$$y+3=\sqrt[3]{2x-1}$$

$$y=\sqrt[3]{2x-1}-3$$

Since $\sqrt[3]{2x-1}$ returns exactly one value for any $x$, the equation $y=\sqrt[3]{2x-1}-3$ defines $y$ as the function of $x$.

12.

$2x-\left|y\right|=0⟹\left|y\right|=2x.$ This equation does not correspond to a function, since for one $x$ we can assign 2 outputs: $x=2⟹y\in \{-4,4\}.$ Therefore, as the vertical line through $x=2$ crosses the graph of this equation in two points, the equation fails the Vertical Line Test.

For a function defined by a table, it suffices to check by hand that no input corresponds to several outputs.

13.

For $x=60$, we have $y\in \{7,8\}.$ Since horizontal line $x=60$ intersects the plot in two points, $(60,7)$ and $(60,8)$, the table below does not represent a function.

14.

Since there is no $x$ for which 2 outputs are assigned, a table determines a function.