Answer:
The function [tex] \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}[/tex] is continuous at x = 36.
Step-by-step explanation:
We need to follow the following steps:
The function is:
[tex] \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}[/tex]
The function is continuous at point x=36 if:
- The function [tex] \\ f(x)[/tex] exists at x=36.
- The limit on both sides of 36 exists.
- The value of the function at x=36 is the same as the value of the limit of the function at x = 36.
Therefore:
The value of the function at x = 36 is:
[tex] \\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}[/tex]
[tex] \\ f(36) = \frac{36*6}{900} = \frac{6}{25}[/tex]
The limit of the [tex] \\ f(x)[/tex] is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.
For this case:
[tex] \\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}[/tex]
[tex] \\ \lim_{x \to 36} f(x) = \frac{6}{25} [/tex]
Since
[tex] \\ f(36) = \frac{6}{25}[/tex]
And
[tex] \\ \lim_{x \to 36} f(x) = \frac{6}{25} [/tex]
Then, the function [tex] \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}[/tex] is continuous at x = 36.
