To check Example 4, Kara graphs y1=(7 x^2+21 x)/(14 x) and y2=(x+3)/(2) since the graphs of y1

step 1 According to given question, we have y1 is equal to 7x squared plus 21x divided by 14x and y2 is

To check Example 4, Kara graphs y1=(7 x^2+21 x)/(14 x) and ...

To check Example $4,$ Kara graphs $$ y_{1}=\frac{7 x^{2}+21 x}{14 x} \text { and } y_{2}=\frac{x+3}{2} $$ since the graphs of $y_{1}$ and $y_{2}$ appear to be identical, Kara believes that the domains of the functions described by $y_{1}$ and $y_{2}$ are the same, $\mathbb{R} .$ How could you convince Kara otherwise?

To check Example $4,$ Kara graphs
$$
y_{1}=\frac{7 x^{2}+21 x}{14 x} \text { and } y_{2}=\frac{x+3}{2}
$$
since the graphs of $y_{1}$ and $y_{2}$ appear to be identical, Kara believes that the domains of the functions described by $y_{1}$ and $y_{2}$ are the same, $\mathbb{R} .$ How could you convince Kara otherwise?

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Key Concepts

Domain of a Rational Functions

The domain of a rational function is the set of all real numbers for which the function is defined. This requires that the denominator of the rational expression does not equal zero, as division by zero is undefined. By identifying these restrictions before simplification, one can accurately represent the function's domain.

Simplification and Cancellation in Rational Expressions

Simplification of rational expressions often involves canceling common factors in the numerator and denominator. However, such cancellation is only valid if the factor being canceled does not equal zero. Even if the simplified form appears to have a broader domain, it is important to remember that the original expression may have domain restrictions that are not apparent after cancellation.

Removable Discontinuities

A removable discontinuity occurs when a function is undefined at a certain point due to a common factor in the numerator and denominator that can be canceled. Although the simplified expression does not show a discontinuity, the original function has a 'hole' at that point, meaning the domain is actually restricted at that value.

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