Define a function f(x)= {(x^2-9)/(x-3), if x≠3,
{c, if x=3
What value of c makes f continuous at x=3?
Is the answer 1?
f(x)=
{(x^2-9)/(x-3), if x≠3,
{c, if x=3
Note that if x≠3, f(x) = x+3
so, you want f(3) = 6
Ohh I see, I've over-thought this question. Thanks for the help!
To determine the value of c that makes the function f(x) continuous at x=3, we need to evaluate the limit of f(x) as x approaches 3 from both sides.
First, let's find the limit of f(x) as x approaches 3 from the left-hand side (x < 3):
lim (x→3-) f(x) = lim (x→3-) [(x^2-9)/(x-3)]
To find this limit, we can substitute x=3 into the function:
lim (x→3-) f(x) = [(3^2-9)/(3-3)] = [(9-9)/(0)]
However, we encounter an indeterminate form of 0/0 here. To resolve this, we can factorize the numerator as a difference of squares:
lim (x→3-) f(x) = [(x-3)(x+3)/(x-3)]
Now, we can cancel out the common factor of (x-3):
lim (x→3-) f(x) = lim (x→3-) (x+3) = 6
Next, let's find the limit of f(x) as x approaches 3 from the right-hand side (x > 3):
lim (x→3+) f(x) = lim (x→3+) [(x^2-9)/(x-3)]
Again, we substitute x=3 into the function:
lim (x→3+) f(x) = [(3^2-9)/(3-3)] = [(9-9)/(0)] = 0/0 (indeterminate form)
Similar to before, we factorize the numerator:
lim (x→3+) f(x) = [(x-3)(x+3)/(x-3)]
Canceling out the common factor:
lim (x→3+) f(x) = lim (x→3+) (x+3) = 6
To make f(x) continuous at x=3, the left-hand limit and right-hand limit need to be equal. In this case, the limits are both equal to 6.
Therefore, the value of c that makes f(x) continuous at x=3 is c=6, not 1.