Question Statement
Q2. f(x)={kx+1,2−kx,x≤3x>0 is continuous at 3. What is k ?
Background and Explanation
For a function to be continuous at a point x=a, the left-hand limit must equal the right-hand limit, and both must equal the function value at that point. When working with piecewise functions defined differently on either side of a boundary, we evaluate each piece separately at the approach point.
Solution
For f(x) to be continuous at x=3, the left-hand limit must equal the right-hand limit:
limx→3−f(x)=limx→3+f(x)
Evaluating the left-hand limit:
For x≤3, we use the expression f(x)=kx+1:
limx→3−f(x)=3k+1
Evaluating the right-hand limit:
For x>0 (specifically in the neighborhood x>3 when approaching from the right), we use the expression f(x)=2−kx:
limx→3+f(x)=2−3k
Setting the limits equal and solving:
3k+1=2−3k
3k+3k=2−1
6k=1
k=61
- Continuity condition: limx→a−f(x)=limx→a+f(x)=f(a)
- Left-hand limit evaluation: Substituting x=3 into kx+1
- Right-hand limit evaluation: Substituting x=3 into 2−kx
- Linear equation solving: Isolating the variable k
Summary of Steps
- Find left-hand limit: Use f(x)=kx+1 for x≤3 to get limx→3−f(x)=3k+1
- Find right-hand limit: Use f(x)=2−kx for x>3 to get limx→3+f(x)=2−3k
- Apply continuity condition: Set 3k+1=2−3k
- Solve: Rearrange to get 6k=1, therefore k=61