For what value of a is
f(x)=(x^2)-a x<3 f(x)=2ax x(> or =)3
continuous at every x?
correction its f(x)=(x^2)-1 not f(x)=(x^2)-a
To be continuous, the two graph must be "linked" at x = 3
that is,
x^2 - 1 = 2ax for x = 3
9-1 = 6a
a = 8/6 = 4/3
so the two functions are
f(x) = x^2 - 1 for x < 3 and f(x) = (8/3)x for x≥ 3
(note that (3,8) would be on both graphs, ignoring the restrictions)
To find the value of "a" for which the function f(x) is continuous at every x, we need to ensure that the function is continuous at the point x = 3.
For a function to be continuous, it must satisfy three conditions:
1. The function f(x) must be defined at x = 3.
2. The limit of f(x) as x approaches 3 from the left must be equal to the limit of f(x) as x approaches 3 from the right.
3. The limit of f(x) as x approaches 3 (from either direction) must be equal to the value of f(3).
Let's evaluate each condition step by step:
1. The function f(x) must be defined at x = 3:
From the given conditions, if x < 3, then f(x) = (x^2) - ax. So at x = 3, we can substitute it into this equation:
f(3) = (3^2) - a(3) = 9 - 3a
2. The limit of f(x) as x approaches 3 from the left must be equal to the limit of f(x) as x approaches 3 from the right:
For x approaching 3 from the left, f(x) = (x^2) - ax
Taking the limit as x approaches 3 from the left side:
lim(x->3-) (x^2 - ax) = 3^2 - 3a = 9 - 3a
For x approaching 3 from the right, f(x) = 2ax
Taking the limit as x approaches 3 from the right side:
lim(x->3+) (2ax) = 2a(3) = 6a
Since the limits from both directions must be equal,
9 - 3a = 6a
3. The limit of f(x) as x approaches 3 must be equal to the value of f(3):
From the above condition, we already found that the limit is equal to 9 - 3a.
So, we need to equate it to f(3):
9 - 3a = f(3) = 9 - 3a
Now, solving this equation:
9 - 3a = 9 - 3a
From this equation, we see that -3a terms cancel each other out. This means that "a" can have any value and the function will always be continuous at x = 3.
Therefore, any value of "a" will make the function f(x) continuous at every x.