F(x)=1/(x^2-5): find largest A so f(x) defined on [-10,A]

[jemadd2]

Suppose F(x)=1/(x^2-5). What is the largest value of A such that f(x) is defined on the interval [-10,A)?

I can't believe I don't know how to do this =/

 

[Deleted member 4993]

Suppose F(x)=1/(x^2-5). What is the largest value of A such that f(x) is defined on the interval [-10,A)?

What is the range of the given function F(x)?

Please share with us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[stapel]

Suppose F(x)=1/(x^2-5). What is the largest value of A such that f(x) is defined on the interval [-10,A)?

Can you divide by zero?

What is the first (leftmost) x-value which would cause division by zero?

:wink:

 

[galactus]

If I am interpreting correctly, $\displaystyle F(x)=\frac{1}{x^{2}-5}$

Then, $\displaystyle F'(x)=f(x)=\frac{-2x}{(x^{2}-5)^{2}}$

Now, what results in division by 0?. That's a start.

 

[BigGlenntheHeavy]

Domain of $\displaystyle f(x): (-\infty,-\sqrt{5})\cup(-\sqrt{5},\sqrt{5})\cup(\sqrt{5},\infty)$. Can you take it from here?

Perhaps I should have said the range of $\displaystyle f(x): \ (-\infty,-.2]\cup(0,\infty)$