Show that f is continuous on (-infinity, infinity).
f(x) = x^2 if x<1 and the square root of x if x is greater than or equal to 1.
I'm not entirely sure how to start this. Help is greatly appreciated :)
f(1) = 1
For x<1, the function approaches 1 as x->1 from below.
Therefore the function is continuous at x=1. It is also defined and continuous for x>1 and x<1.
Therefore it is continuous on (-infinity, infinity)
Many thanks!
To show that the function f(x) is continuous on (-∞, ∞), we need to prove that it is continuous at every point within this interval.
First, let's consider the interval (-∞, 1).
Within this interval, the function is defined as f(x) = x^2 for all x < 1.
For any value of x in (-∞, 1), f(x) satisfies the equation f(x) = x^2, which is a polynomial function. Polynomial functions are continuous for all real numbers. Therefore, f(x) is continuous within the interval (-∞, 1).
Next, let's examine the interval [1, ∞).
Within this interval, the function is defined as f(x) = √x for all x ≥ 1.
For any value of x in [1, ∞), f(x) satisfies the equation f(x) = √x, which is a square root function. Square root functions are continuous for all x ≥ 0. Since x ≥ 1 within this interval, the condition for continuity is met. Therefore, f(x) is continuous within the interval [1, ∞).
Now, let's consider the point x = 1, where the two intervals meet.
To prove continuity at x = 1, we need to show that the limit of f(x) as x approaches 1 from both sides is equal to the value of f(1).
First, let's calculate the limit as x approaches 1 from the left side:
lim(x→1-) f(x) = lim(x→1-) x^2 = 1^2 = 1.
Next, let's calculate the limit as x approaches 1 from the right side:
lim(x→1+) f(x) = lim(x→1+) √x = √1 = 1.
Since the limit from both the left and right sides is equal to 1, which is the value of f(1), the limit of f(x) as x approaches 1 exists.
Therefore, f(x) is continuous at x = 1.
Since f(x) is continuous on all intervals (-∞, 1), [1, ∞), and at x = 1, it can be concluded that f(x) is continuous on the interval (-∞, ∞).
To show that a function is continuous on a given interval, we need to demonstrate three things:
1. The function is defined at every point within the interval.
2. The limit of the function exists as x approaches any point within the interval.
3. The value of the function at each point in the interval is equal to the limit at that point.
Let's apply these principles to the given function, f(x) = x^2 if x < 1 and the square root of x if x is greater than or equal to 1.
1. Check for the definition of the function at all points in the interval:
The given function is defined for all values of x, including both x < 1 and x ≥ 1. Therefore, it is defined at every point within the interval (-∞, ∞).
2. Compute the limits of the function as x approaches any point within the interval:
a) Let's first consider the limit as x approaches a point within x < 1 (left-hand limit):
lim(x→a-, x^2) = a^2, for any real number a < 1.
b) Now, let's evaluate the limit as x approaches a point within x ≥ 1 (right-hand limit):
lim(x→a+, √x) = √a, for any real number a ≥ 1.
Both the left-hand and right-hand limits exist for every point within the interval (-∞, ∞), as long as a is a real number.
3. Compare the value of the function at each point with the corresponding limit:
a) For x < 1: f(x) = x^2, and lim(x→a-, x^2) = a^2.
The values of the function and the limit are the same for every point x < 1.
b) For x ≥ 1: f(x) = √x, and lim(x→a+, √x) = √a.
Again, the values of the function and the limit are the same for every point x ≥ 1.
Since all three conditions are satisfied for the function f(x) = x^2 if x < 1 and √x if x is greater than or equal to 1, we can conclude that the function is continuous on the interval (-∞, ∞).