Test or construct a smooth piecewise join

Two straight branches meet at the same point but have different gradients. Is the joined function differentiable there?

Change the right gradient through all seven settings. The branch values remain equal

Which value of m makes this join differentiable?

Two comparisons, in order

The symbols L and R name the branch polynomials. First evaluate L(a) and R(a). If they differ, stop: continuity fails. If they agree, compare L′(a) and R′(a)

1. 1Check continuity first Evaluate both branch polynomials at the join. With one branch including the join, equal branch values give continuity. A jump prevents differentiability even if the two branch gradients agree
2. 2A corner is continuous but not smooth Once continuity holds, differentiate both branches and compare their values at the join. Equal finite one-sided gradients give differentiability for these polynomial branches \(\displaystyle L(a)=R(a),\quad L'(a)=R'(a)\)

Given

\(\displaystyle L(x)=x^2+2x,\quad a=1\)

Evaluate the left branch at the join x = 1.

Given

\(\displaystyle R(x)=4x-1,\quad a=1\)

Evaluate the right branch at x = 1.

Given

\(\displaystyle L(x)=x^2+2x,\quad L'(1)\)

Find the left branch gradient at x = 1.

The right rule is R(x) = 4x − 1. Both branch values at 1 equal 3 and the left gradient equals 4. What follows?

A derivative match cannot repair a jump

For L(x) = 2x + 1 and R(x) = 2x + 4 at x = 0, both branch gradients are 2, but the values are 1 and 4. The piecewise function is not differentiable. At a jump, L′(a) and R′(a) are derivatives of the branch extensions, not necessarily one-sided derivatives of the piecewise function

1. 1Check continuity first Evaluate both branch polynomials at the join. With one branch including the join, equal branch values give continuity. A jump prevents differentiability even if the two branch gradients agree

A piecewise rule includes its left branch at the join. Is that endpoint convention enough to guarantee continuity?

Construct a smooth join

Consider x² + 2x for x ≤ 1 and ux + v for x > 1. The unknown constants must satisfy both the value equation u + v = 3 and the gradient equation u = 4.

1. 1Solve both conditions for parameters Use value matching and gradient matching as two simultaneous equations. Check the recovered parameters in both equations, rather than stopping after continuity

Given

\(\displaystyle u=2(1)+2\)

Use the gradient equation to determine u.

Given

\(\displaystyle u+v=3,\quad u=4\)

Use continuity after finding u = 4 to determine v.

The left branch is x² + 2x for x ≤ 1, and the right branch is ux + v for x > 1. A student chooses u = 2 and v = 1. The branch values at 1 both equal 3. Is the join smooth?

Unknowns can be on either side

The same checks apply when the affine branch is on the left, or when one unknown is a constant in a quadratic and the other is a linear coefficient. Keep the function input separate from the parameter names

1. 1Solve both conditions for parameters Use value matching and gradient matching as two simultaneous equations. Check the recovered parameters in both equations, rather than stopping after continuity

Given

\(\displaystyle 2^2+u=4(2)-2\)

Find u for the split-parameter example once v = 4.

More than one join

Evaluate the value and gradient conditions at each join separately. Two continuous corners remove two inputs from the derivative domain. A zero gradient within a smooth branch still exists

1. 1Check every join separately A piecewise function can have more than one join. Its derivative exists within each polynomial branch, but each join needs its own continuity and gradient checks

A three-branch linear function is continuous at joins −1 and 2. Its successive slopes are 1, −1 and 1. Where does its derivative fail to exist?