Finding removable discontinuity

[i5hands]

Homework Statement

Let f(x)= {x^2-7x+10, for x^2 ≠ 25
{ A, for x = 5
{ B, for x = -5
Is there a value of A that makes f continuous at x= 5?
Is there a value of B that makes f continuous at x= -5?

 

[STEMucator]

Homework Statement

Let f(x)= {x^2-7x+10, for x^2 ≠ 25
{ A, for x = 5
{ B, for x = -5
Is there a value of A that makes f continuous at x= 5?
Is there a value of B that makes f continuous at x= -5?

You should draw out the graph of f, it will be easier to see.

So f(x) = x²-7x+10 when x≠±5. So f(x) is a parabola at every point except when x=5 or x=-5.

At those two points it attains a value A for x=5 and B for x=-5 respectively. Are there values of A and B you can choose to make the graph smooth? As in have no jumps or breaks in it?

Big hint : What is f(5) and f(-5) ? That should tell you something about the roots of f and the values you need.

 

[i5hands]

Thank you!
There is a Vertical asymptote at x=-5 so does that mean there are no values to make it continuos?

 

[i5hands]

i made a mistake when writing the question, it is f(x) ={ x^2-7x+10 / x^2 - 25

 

[STEMucator]

Ah even better then.

Can you factor : x^2-7x+10 ?
Can you factor : x^2-25?

Now simplify your f(x) after factoring, what do you get and what do you notice?

 

[i5hands]

Yes, you can factor so there is a value for a when x = 5 you get 3/10. But there is not a value for when x=-5 because after plugging in -5 you get undefined therefore there are no values for B.
Thank you so much!

 

[STEMucator]

No problem bud :)