Suppose f of x equals 3 times x minus k when x is less than 5 and equals 1 plus k times x when x is greater than or equal to 5.Find the value of k that would make f continuous at x = 5.
-3
0
7/3
no such k will make f continuous
suppose you use symbols instead of words:
f(x)
= 3x-k for x < 5
= kx+1 for x >= 5
for f to be continuous at x=5, we need
15-k = 5k+1
14 = 6k
k = 7/3
To find the value of k that would make f continuous at x = 5, we need to check the two different expressions for f at x = 5 and make sure they give us the same value.
The given function f(x) is defined as follows:
f(x) = 3x - k, when x < 5
f(x) = 1 + kx, when x ≥ 5
To find the value of f at x = 5, we evaluate both expressions:
f(5) = 3(5) - k, using the first expression
f(5) = 15 - k
f(5) = 1 + k(5), using the second expression
f(5) = 1 + 5k
To make f continuous at x = 5, the values obtained from both expressions should be equal.
Therefore, we set the two expressions equal to each other and solve for k:
15 - k = 1 + 5k
Simplifying the equation:
-6k = -14
k = -14 / -6
k = 7/3
So the value of k that would make f continuous at x = 5 is 7/3.