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
***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.
To determine what the function shows at x=5, we need to evaluate the function at x=5.
Given the function:
f(x) = (x^2 - 25)/(x - 5)
To evaluate the function at x=5, we substitute x=5 into the function:
f(5) = (5^2 - 25)/(5 - 5)
Simplifying this expression:
f(5) = (25 - 25)/(5 - 5)
f(5) = 0/0
Notice that we get an indeterminate form of 0/0 when evaluating the function at x=5. This indicates a potential discontinuity at that point.
Among the given choices, the type of discontinuity present at x=5 is a removable discontinuity (choice A).
To determine what the function shows at x=5, we need to evaluate the function at that specific value.
Given the function:
f(x) = (x^2 - 25) / (x - 5)
To find the value of the function at x=5, we substitute x=5 into the function and simplify.
f(5) = (5^2 - 25) / (5 - 5)
Simplifying further:
f(5) = (25 - 25) / 0
Here, we encounter division by zero (0). Division by zero is undefined in mathematics, which means the function is undefined at x=5.
Therefore, the answer is E. None of the above, as none of the given options describe an undefined value at x=5.