Find a value of k that makes f(x)= x^2,x is less than or equal to 1
and sin(kx), x > 1
continuous at x=1
at x=1, x^2 = 1
so, you need at x=1, sin(k) = 1
so where is sin(k) equal to 1?
I understand that, but Im not sure how to solve for the k ):
come on sin(π/2) = 1. So, k=π/2
so, f(x) = sin(π/2 x) for x>1
That way the limit on both left and right is f(x) = 1
Thank you.. I forgot about the unit circle
To find the value of k that makes the function f(x) continuous at x = 1, we need to ensure that the left-hand limit of f(x) as x approaches 1 from the left is equal to the right-hand limit of f(x) as x approaches 1 from the right.
Let's start by calculating the left-hand limit, which is the limit of f(x) as x approaches 1 from the left side:
lim(x→1-) f(x)
For x less than or equal to 1, the function is given by f(x) = x^2. Therefore, the left-hand limit can be found by substituting x = 1 into the expression for f(x):
lim(x→1-) (x^2) = (1^2) = 1
Next, let's calculate the right-hand limit, which is the limit of f(x) as x approaches 1 from the right side:
lim(x→1+) f(x)
For x greater than 1, the function is given by f(x) = sin(kx). Substituting x = 1 into the expression for f(x), we get:
lim(x→1+) (sin(kx)) = sin(k)
Now, for the function to be continuous at x = 1, the left-hand and right-hand limits must be equal. Therefore, we have the equation:
1 = sin(k)
To find the value of k, we need to solve this equation. Since sin(k) can take multiple values between -1 and 1, there will be multiple solutions for k. We can use trigonometric properties or a calculator to find the solutions.
For example, if we assume k to be between 0 and 2π, we can find the solutions using a calculator or a trigonometric table. Let's say one of the solutions is k = π/6.
Therefore, the value of k that makes the function f(x) continuous at x = 1 is k = π/6.