2

Case 1:

$$\frac{2x^2}{4x^3}$$

$$\frac{1}{2x}$$

The graph hasn't changed.

Case 2:

$$\frac{6t^4+4t^2}{t}$$

$$6t^3+4t$$

The graph hasn't changed.

Case 3:

$$\frac{x^3-8}{x^2-4}$$

$$\frac{(x-2)(x^2+2x+4)}{(x+2)(x-2)}$$

$$\frac{(x^2+2x+4)}{(x+2)}$$

The graph hasn't changed.

Question(s):

  1. How will I know that the graph will be the same after my operations?
  2. What things should I always avoid if I want to keep the function/graph the same?
Misha Lavrov
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tryingtobeastoic
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1 Answers1

2

For case $1$ we have that the two expressions are indeed equivalent since they are defined in both cases for $x\neq 0$.

For case $2$ the function are different because the first expression is not defined at $t=0$.

For case $3$ the function are different because the first expression is not defined at $x=2$.

More in general for $b(x_0)\neq 0$

$$f(x)=\frac{(x-x_0)a(x)}{(x-x_0)b(x)} \quad g(x)=\frac{a(x)}{b(x)}$$

have the same graph on all the domain of definition but $f(x)$ is not defined at $x=x_0$. The two expressions are equivalent if also $g(x)$ is not defined at $x=x_0$.

Refer also to the related

user
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