Use the definition of continuity and properties of limits to show that the function is continues at given a.
f(x)=x^2+sqrt(7-x) a=4
To show that the function f(x) = x^2 + sqrt(7 - x) is continuous at a = 4, we need to demonstrate that three conditions are met:
1. The function is defined at a = 4.
2. The limit of the function exists as x approaches a = 4.
3. The limit of the function is equal to the value of the function at a = 4.
Let's go through these conditions one by one:
1. The function is defined at a = 4:
To check if the function is defined at x = 4, we substitute 4 into the function:
f(4) = 4^2 + sqrt(7 - 4)
= 16 + sqrt(3)
= 16 + √3
Since this value is defined, the first condition is satisfied.
2. The limit of the function exists as x approaches a = 4:
To find the limit as x approaches 4, we evaluate the left-hand limit and the right-hand limit separately:
Left-hand limit (as x approaches 4 from the left):
Limit[x->4-] (x^2 + sqrt(7 - x))
Let's evaluate this limit. As x approaches 4 from the left, 7 - x approaches 3.
Limit[x->4-] (x^2 + sqrt(7 - x)) = Limit[x->4-] (x^2 + sqrt(3))
Since both terms in the expression are continuous, the limit exists.
Right-hand limit (as x approaches 4 from the right):
Limit[x->4+] (x^2 + sqrt(7 - x))
Let's evaluate this limit. As x approaches 4 from the right, 7 - x approaches 3.
Limit[x->4+] (x^2 + sqrt(7 - x)) = Limit[x->4+] (x^2 + sqrt(3))
Again, since both terms in the expression are continuous, the limit exists.
Since the left-hand limit and the right-hand limit both exist, the limit of the function as x approaches 4 exists.
3. The limit of the function is equal to the value of the function at a = 4:
We have already calculated the value of the function at x = 4 in step 1:
f(4) = 16 + √3
Now, we compare the value of the function at a = 4 with the limit:
Limit[x->4] (x^2 + sqrt(7 - x)) = Limit[x->4] (x^2 + sqrt(3))
= 16 + √3
Since the limit is equal to f(4), the third condition is satisfied.
Since all three conditions (being defined, having a limit, and having the limit equal to the value) are met, we can conclude that the function f(x) = x^2 + sqrt(7 - x) is continuous at x = 4.