what does this function show at x=5?
f(x)=^2-25/x-5
A. removable discountinuity
B. jump discountinuity
C. infinite discountinuity
D. Countinuity
E. None of the above
2 answers
***x^2-25
I will assume you meant
f(x) = (x^2 - 25)/(x-5)
which reduces to f(x) = x+5 , x ≠ 5
so when x = 5
f(x) = 0/0 , which is indeterminate
I am not familiar with terms such as "jump discountinuity" , (did you mean discontinuity ? )
but the graph would be the graph of
y = x+ 5 with a "hole" at the point (5,10)
Use the definitions of the above, which must have been given to you or are found in your text, to apply to the answer I gave you.
f(x) = (x^2 - 25)/(x-5)
which reduces to f(x) = x+5 , x ≠ 5
so when x = 5
f(x) = 0/0 , which is indeterminate
I am not familiar with terms such as "jump discountinuity" , (did you mean discontinuity ? )
but the graph would be the graph of
y = x+ 5 with a "hole" at the point (5,10)
Use the definitions of the above, which must have been given to you or are found in your text, to apply to the answer I gave you.
1 answer