Solving F(x) for x=9: Defining g(x) - Joe

[cowgiljl]

Can anybody help me get started on this the radical in the Numerator gots me lost.

F(x) = 3-sq root x / 9-x has a removable discontinuity at x=9. define a new function g(x) such that g=f everywhere except at x = 9 , but is continuous everywhere.



thanks joe

 

[matt grime]

firstly you ought to bracket things to make it clear what is divided by what, secondly, can you write 9-x as a product of two factors (involving x^{1/2} perhaps, difference of two "squares").

 

[M98Ranger]

Hi Joe,

To solve for F(x) when x=9, we can first substitute 9 into the given function:

F(9) = 3-sqrt(9) / 9-9

We know that the square root of 9 is 3, and 9-9 is equal to 0. This means that the numerator is 3-3, which is equal to 0. And since any number divided by 0 is undefined, this is why we have a removable discontinuity at x=9.

To define a new function g(x) that is continuous everywhere except at x=9, we can consider the following steps:

1. Choose a value for g(9) that will make the function continuous at x=9. Since the numerator becomes 0 when x=9, we can choose any value for g(9) as long as it does not result in a division by 0.

2. Let's choose g(9) = 1. This means that our new function g(x) will be equal to 1 when x=9.

3. Now, we need to define g(x) for all other values of x. Since g(x) should be equal to F(x) everywhere except at x=9, we can simply let g(x) = F(x) for all values of x except x=9.

Therefore, our new function g(x) is defined as:

g(x) = 3-sqrt(x) / 9-x for all x ≠ 9
g(9) = 1

This new function g(x) is continuous everywhere except at x=9, where there is still a removable discontinuity. I hope this helps you understand how we can define a new function to make it continuous at a specific point. Let me know if you have any other questions or need further clarification. Good luck!