let f(x)= kxsquared, if x≤2
10 - kx, if x>2
what value of k makes f continuous at x=2
What feature will the graph of f have at this point?
To find the value of k that makes f continuous at x = 2, we need to make sure that the left-hand limit (as x approaches 2 from the left) is equal to the right-hand limit (as x approaches 2 from the right).
First, let's find the left-hand limit of f as x approaches 2 from the left:
lim(x→2-) f(x) = lim(x→2-) kx^2
Since x approaches 2 from the left, we can replace x with 2 to evaluate the limit:
lim(x→2-) f(x) = k(2)^2 = 4k
Next, let's find the right-hand limit of f as x approaches 2 from the right:
lim(x→2+) f(x) = lim(x→2+) (10 - kx)
Again, we can replace x with 2 because x approaches 2 from the right:
lim(x→2+) f(x) = 10 - k(2) = 10 - 2k
Now, we set the left-hand limit equal to the right-hand limit:
4k = 10 - 2k
Simplifying the equation, we get:
6k = 10
Dividing both sides by 6, we find:
k = 10/6 = 5/3
Therefore, the value of k that makes f continuous at x = 2 is k = 5/3.
Now, let's consider the graph of f at x = 2. Since f(x) changes its definition at x = 2, we can expect a discontinuity in the graph. The left-hand side of the graph will be a parabola, while the right-hand side will be a line. At x = 2, the two portions of the graph will meet and create a corner-like point, which is a type of discontinuity known as a "corner point" or a "cusp."