Determine whether the function f(x) is continuous at x=-3. f(x)={[15-x^(2)",",x = 3]:} f(x) is continuous at x=-3 f(x) is discontinuous at x=-3

Question

Determine whether the function  f(x) is continuous at  x=-3.  f(x)={[15-x^(2)",",x < 3],[15-2x",",x >= 3]:}  f(x) is continuous at  x=-3  f(x) is discontinuous at  x=-3

Bytelearn · Accepted Answer

Check Conditions: To determine if the function f(x) is continuous at x=-3, we need to check if the following three conditions are met: 1. f(-3) is defined. 2. The limit of f(x) as x approaches -3 exists. 3. The limit of f(x) as x approaches -3 is equal to f(-3).Find f(-3): First, we need to find f(-3) using the appropriate piece of the function. Since -3 is less than 3, we use the first piece of the function, which is 15 - x^2. f(-3) = 15 - (-3)^2 = 15 - 9 = 6.Find Limit Left: Next, we need to find the limit of f(x) as x approaches -3 from the left (x -> -3^-) and from the right (x -> -3^+). For x -> -3^-, we use the first piece of the function, 15 - x^2. lim (x -> -3^-) f(x) = lim (x -> -3^-) (15 - x^2) = 15 - (-3)^2 = 15 - 9 = 6.Find Limit Right: For x -> -3^+, we still use the first piece of the function, 15 - x^2, because the change in the function's definition occurs at x = 3, not at x = -3. lim (x -> -3^+) f(x) = lim (x -> -3^+) (15 - x^2) = 15 - (-3)^2 = 15 - 9 = 6.Verify Continuity: Since the limit from the left and the limit from the right are both equal to 6, and f(-3) is also 6, all three conditions for continuity are satisfied. Therefore, f(x) is continuous at x = -3.

Determine whether the function $f(x)$ is continuous at $x=-3$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 15-x^{2}, & x<3 \\ 15-2 x, & x \geq 3 \end{array}\right.$ $\newline$ $f(x)$ is continuous at $x=-3$ $\newline$ $f(x)$ is discontinuous at $x=-3$

Full solution

More problems from Find derivatives of logarithmic functions

Related topics