Let f(x)=(x^-9)/x-3 where x cannot equal 3
and f(x)=2x where x =3
Determine is f(x) is continuous at x=3
How do we know?
denominator of x - 3 makes the function undefined at x = 3
UNLESS you have a typo and your numerator is supposed to be
(x^2-9)
which is (x-3)(x+3)
if that is the case the (x-3) cancels
and you really have
f(x) = x+3
nw when x = 3, that is 6
THE SAME as 3 x = 2*3 = 6
so yes
I mean 2x= 2*3 = 6 of course
Would this be a good answer? Yes this function is continuous because the point value is defined which is f(3). The limit as x approaches 3 exists from both left and right and the point value f(3) is equal to the limit
To determine if a function is continuous at a particular point, we need to check three conditions:
1. The function should be defined at that point.
2. The left-hand limit should be equal to the right-hand limit.
3. The function value at the point should be equal to the limit.
Let's check these conditions for the function f(x) given:
1. At x = 3, the function is defined as f(x) = 2x, so the function is defined at x = 3.
2. To find the left-hand limit, we need to evaluate the function as x approaches 3 from the left side (x < 3). Let's calculate the limit:
lim(x→3-) (x^-9)/(x-3)
= lim(x→3-) (1/x^9)/(1/x)
= lim(x→3-) (1/x^8)
As x approaches 3 from the left side, (1/x^8) approaches (1/3^8).
3. To find the right-hand limit, we need to evaluate the function as x approaches 3 from the right side (x > 3). Let's calculate the limit:
lim(x→3+) 2x = 2 * 3 = 6
Now, let's compare the limits:
lim(x→3-) (1/x^8) = 1/3^8 = 1/6561
lim(x→3+) 2x = 6
The left-hand limit is not equal to the right-hand limit. Hence, the function f(x) is not continuous at x = 3.