Let f (x) be defined by: f (x) = 2x /（3 + x） , x ≤ −2 k(x − 2), −2 < x < 4 2x − 6, x ≥ 4 Which of the following values of k would make f (x) continuous on R? a) k = 2 (b) k = 1 (c) There is no such value for k. (d) k = −1
Question
Let f (x) be defined by: f (x) = 2x /（3 + x） , x ≤ −2 k(x − 2), −2 < x < 4 2x − 6, x ≥ 4 Which of the following values of k would make f (x) continuous on R? a) k = 2 (b) k = 1 (c) There is no such value for k. (d) k = −1
Solution
For a function to be continuous, the limit from the left and the right at each point must be equal.
Let's find the limit of each piece of the function at the points of interest, which are x = -2 and x = 4.
-
At x = -2:
From the left: lim (x->-2-) f(x) = 2*(-2) / (3 - 2) = -4
From the right: lim (x->-2+) f(x) = k*(-2 - 2) = -4k
For these to be equal, -4 = -4k, so k = 1.
-
At x = 4:
From the left: lim (x->4-) f(x) = k*(4 - 2) = 2k
From the right: lim (x->4+) f(x) = 2*4 - 6 = 2
For these to be equal, 2k = 2, so k = 1.
Therefore, the value of k that makes f(x) continuous on R is k = 1, so the answer is (b) k = 1.
Similar Questions
Determine the value of ‘k’ for which the following function is continuous at x = 3: (All India 2017)
Find the value for k so that the function will be continuous at x=4𝑥=4.f(x)=−8x2+64x−128x−4−73x2+24x+48+kifx<4ifx≥4
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.Tutorial ExerciseFor what value of the constant c is the function f continuous on (−∞, ∞)?f(x) = cx2 + 5x if x < 6x3 − cx if x ≥ 6Part 1 of 2Note that f is continuous on (−∞, 6) and (6, ∞). For the function to be continuous on (−∞, ∞), we need to ensure that as x approaches 6, the left and right limits match.First we find the left limit.lim x→6− f(x) = lim x→6− (cx2 + 5x) = Next, we find the right limit.lim x→6+ f(x) = lim x→6+ (x3 − cx) =
Which of the following functions is continuous for every value of x except x=0?
The table below shows certain values of k(x) on the interval [–17,–5].x k(x)–17 –11–13 –5–9 10–5 –14If k(x) is continuous on [–17,–5], can you conclude that the equation k(x)=–4 has a solution?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.