Chapter 1.8, Problem 50E

To determine

To calculate: Whether the set of points $\left(1,1\right),\left(1,\frac{1}{2}\right)\text{ and }\left(-1,\frac{1}{2}\right)$ lie on the graph of the equation $y\left(x^2+1\right)=1$

Answer to Problem 50E

The points $\left(1,\frac{1}{2}\right)\text{ and }\left(-1,\frac{1}{2}\right)$ lie on the graph of the equation whereas the point $\left(1,1\right)$ does not lie on the graph.

Explanation of Solution

Given information:

The set of points $\left(1,1\right),\left(1,\frac{1}{2}\right)\text{ and }\left(-1,\frac{1}{2}\right)$ and equation of graph $y\left(x^2+1\right)=1$ .

Formula used:

The set of all points $\left(x,y\right)$ in the Cartesian plane or coordinate plane that satisfy the equation is known as the graph of the equation in x and y coordinates.

Calculation:

Consider the providedset of points $\left(1,1\right),\left(1,\frac{1}{2}\right)\text{ and }\left(-1,\frac{1}{2}\right)$ and equation of graph $y\left(x^2+1\right)=1$ .

Recall that the set of all points $\left(x,y\right)$ in the Cartesian plane or coordinate plane that satisfy the equation is known as the graph of the equation in x and y coordinates.

Substitute the point $\left(1,1\right)$ in the equation $y\left(x^2+1\right)=1$ and check whether left hand side is equal to right hand side.

  $\begin{array}{c}1\left({\left(1\right)}^2+1\right)=1\\ 2=1\end{array}$

The above equality does not hold so, the point $\left(1,1\right)$ does not lie on the graph of the equation.

Substitute the point $\left(1,\frac{1}{2}\right)$ in the equation $y\left(x^2+1\right)=1$ and check whether left hand side is equal to right hand side.

  $\begin{array}{c}\frac{1}{2}\left({\left(1\right)}^2+1\right)=1\\ \frac{1}{2}\left(2\right)=1\\ 1=1\end{array}$

The above equality holds true so, the point $\left(1,\frac{1}{2}\right)$ lie on the graph of the equation.

Substitute the point $\left(-1,\frac{1}{2}\right)$ in the equation $y\left(x^2+1\right)=1$ and check whether left hand side is equal to right hand side.

  $\begin{array}{c}\frac{1}{2}\left({\left(-1\right)}^2+1\right)=1\\ \frac{1}{2}\left(2\right)=1\\ 1=1\end{array}$

The above equality holds true so, the point $\left(-1,\frac{1}{2}\right)$ lie on the graph of the equation.

Thus, the points $\left(1,\frac{1}{2}\right)\text{ and }\left(-1,\frac{1}{2}\right)$ lie on the graph of the equation whereas the point $\left(1,1\right)$ does not lie on the graph.