Check continuity of piecewise functions at boundary points
Define f(x) with two pieces: { a1*x+b1 for x<c, a2*x+b2 for x>=c }
f(x)={x+, x<
f(x)={x+, x>=2
Result
Left Limit
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Right Limit
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f(c)
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Derivation
Piecewise Continuity Test
1. Compute f(c) = a2*c+b2 (second piece)
2. Left limit: a1*c+b1 (first piece)
3. Right limit: a2*c+b2 (second piece)
4. Equal? Continuous at boundary
Checking continuity of piecewise functions requires evaluating the function and both one-sided limits at each boundary point. The function is continuous at the boundary if the left limit, function value, and right limit are all equal.
⚠The function value at the boundary comes from the piece that includes the equality (x>=c). Adjust constants to make boundary continuous.
What Is Piecewise Continuity?
A piecewise function is continuous if each piece is continuous on its interval AND the function matches at the boundaries. At each boundary, the left piece value, function value, and right piece value must all be equal. If not, there is a jump or removable discontinuity.
Continuity Conditions
lim(x->c-)=lim(x->c+)=f(c). All three must be equal. f(c) defined by the piece that includes c.
Jump Discontinuity
Left and right limits are finite but not equal. The function jumps from one value to another at the boundary.
Removable (Fixing)
If left=right but f(c) differs, the discontinuity can be removed by redefining f(c) to match the limit.
Adjusting Parameters
To make continuous: set a1*c+b1 = a2*c+b2. Solve for the unknown constant. Choose b2 to match at boundary.
Teaching Example: f(x)={2x, x<2; 3x-2, x>=2}. At x=2: left=2*2=4, right=3*2-2=4, f(2)=3*2-2=4. All equal -> CONTINUOUS! Check: 2*2=4 and 3*2-2=4. Both pieces meet at (2,4).
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