Parameters at a Join
A piecewise formula can use different expressions on either side of a point. Even when each expression behaves smoothly on its own, the two pieces may fail to meet properly at the join. Choosing parameters to make the function differentiable requires checking what happens to the difference quotient from both sides.
There are two related but distinct checks. First, the function must be continuous at the join. Second, the left and right difference quotient limits must exist as finite numbers and agree. Continuity is necessary for differentiability, but continuity alone does not make the one-sided slopes equal.
For a piecewise function, the value \(f(a)\) in both quotients is the actual value assigned at the join. It is not automatically the value suggested by either neighboring formula. Keeping this value in the calculation is essential: parameter conditions may be needed just to make the quotient limits finite.
Why Continuity Is a Necessary Check
If the difference quotient has a finite limit, then the function is continuous at the point. To see why, write the change in function value as the product of the input change and the difference quotient.
As \(h\to0\), the first factor tends to zero and, if \(f'(a)\) exists, the quotient tends to the finite number \(f'(a)\). Their product therefore tends to zero. Thus \(f(a+h)\to f(a)\), which is continuity at \(a\).
This gives a useful order for solving parameter problems: calculate the one-sided quotients, impose the conditions that make them finite, and then require their limits to agree. Also check continuity directly by comparing the one-sided function limits with \(f(a)\). The continuity check confirms that the function values meet; it does not replace the derivative calculation.
A General Setup
Suppose the function is given by a left formula \(L(x)\) for \(x<a\), a specified value \(f(a)=c\), and a right formula \(R(x)\) for \(x>a\). The two relevant quotients are
For polynomial expressions, if the numerator does not vanish at \(x=a\), dividing by \(x-a\) makes the quotient unbounded as \(x\) approaches \(a\). A finite one-sided limit therefore requires \(L(a)=c\) on the left and \(R(a)=c\) on the right. After imposing those conditions, factor out \(x-a\) and evaluate the remaining factor at \(a\). Finally, equate the left and right limits.
In parameter problems, these conditions may determine every parameter, leave one or more parameters free, or be inconsistent. The calculation—not an assumption that a unique answer must exist—decides which outcome occurs.
Worked Example: Determine All Three Parameters
Worked Example: Determine All Three Parameters
For the function
find \(a,b,c\) so that \(f\) is differentiable at \(x=1\).
The difference quotient uses \(f(1)=b\). From the left,
For a finite limit, the numerator must be zero at \(x=1\). This gives \(1+a-b=0\), so \(b=1+a\). Substituting that relation and factoring,
The left quotient is \(x+1+a\), so its limit is \(2+a\).
From the right,
A finite limit requires \(3+c-b=0\), or \(b=3+c\). With this relation, the numerator is \(3(x-1)\), so the right quotient is \(3\). Equality of the one-sided limits requires \(2+a=3\), hence \(a=1\). Then \(b=1+a=2\), and \(c=b-3=-1\).
Check the resulting function. At \(x=1\), the left formula gives \(1^2+1(1)=2\), the assigned value is \(2\), and the right formula gives \(3(1)-1=2\). Thus the function is continuous there. The left quotient simplifies to \(x+2\), whose limit is \(3\); the right quotient is \(3\), whose limit is also \(3\). Therefore the function is differentiable at \(1\), with \(f'(1)=3\).
Worked Example: Continuity Does Not Set the Slopes
Worked Example: Continuity Does Not Set the Slopes
Consider the parameterized function
Find the conditions for continuity at zero and, separately, for differentiability there.
The left-hand function limit is \(0\), since \(\alpha x+x^2\to0\) as \(x\to0^-\). The right-hand function limit is also \(0\), since \(\gamma x+x^2\to0\) as \(x\to0^+\). Continuity requires both limits to equal \(f(0)=\beta\). Thus continuity holds exactly when \(\beta=0\); it places no restriction on \(\alpha\) or \(\gamma\).
Now calculate the difference quotients using the assigned value \(\beta\):
If \(\beta\ne0\), the term \(-\beta/x\) has no finite limit at zero. Hence finite one-sided derivative limits require \(\beta=0\), agreeing with the continuity condition. Once \(\beta=0\), the left quotient tends to \(\alpha\), and the right quotient tends to \(\gamma\). They agree exactly when \(\alpha=\gamma\).
Therefore the function is differentiable at zero exactly when \(\beta=0\) and \(\alpha=\gamma\). For a concrete check, choose \(\alpha=2\), \(\beta=0\), and \(\gamma=-1\). Both function limits equal \(f(0)=0\), so the function is continuous. But the one-sided quotient limits are \(2\) and \(-1\), so the function is not differentiable at zero. This verifies that continuity alone does not ensure matching derivatives.
Worked Example: A Family of Valid Choices
Worked Example: A Family of Valid Choices
At the join \(x=2\), consider
Find all parameter relationships that make \(f\) differentiable at \(2\).
The left quotient is
Its numerator must vanish at \(2\), which gives \(4+2p-q=0\), or \(q=4+2p\). Under this condition,
The left quotient is therefore \(x+2+p\), with limit \(4+p\).
The right quotient is
Its numerator must vanish at \(2\), so \(4r+s-q=0\), or \(q=4r+s\). Using that relation gives
The right quotient is \(r(x+2)\), with limit \(4r\). Matching the limits gives \(4+p=4r\), so \(p=4r-4\). Substitution into the value conditions yields
Thus all differentiable choices are described by \(r\) free, with \(p=4r-4\), \(q=8r-4\), and \(s=4r-4\). The conditions do not determine a unique set of parameters.
For an explicit check, take \(r=2\). Then \(p=4\), \(q=12\), and \(s=4\). The function values at the join match: the left expression gives \(2^2+4(2)=12\), the assigned value is \(12\), and the right expression gives \(2(2^2)+4=12\). The left quotient factors to \(x+6\), with limit \(8\); the right quotient factors to \(2(x+2)\), also with limit \(8\). The selected parameters therefore give differentiability at \(2\).
A Common Mistake
A frequent error is to differentiate each formula on its own and compare the resulting slopes, without checking the value assigned at the join. This can produce a false conclusion because the difference quotient subtracts \(f(a)\), not an arbitrary endpoint value from one branch.
For instance, use the function in the second example with \(\alpha=\gamma=0\) and \(\beta=1\). The derivatives of the two branch formulas, considered away from zero, both tend to zero as their inputs approach zero. But the actual difference quotients are, on both sides,
This expression has no finite limit as \(x\to0\), and the function is not even continuous there: the two branch limits are \(0\), while \(f(0)=1\). The correction is to use the actual join value in each difference quotient and check continuity rather than infer differentiability from branch slopes alone.
A Reliable Solving Procedure
Use the left formula in the left quotient and the right formula in the right quotient; use the assigned value \(f(a)\) in both.
For polynomial numerators over \(x-a\), set each numerator equal to zero at \(a\), then factor out \(x-a\).
Equate the resulting left and right limits and solve the parameter equations.
Compare both branch limits with \(f(a)\), then substitute the parameter values into the quotients and confirm their limits agree.
This approach separates the value-matching conditions from the slope-matching condition. It also makes clear whether the constraints give one solution, a family, or no possible choice.
Check Your Understanding
For each question, begin with the difference quotient at the join and keep the assigned function value explicit.
- For a piecewise function with join \(a\) and \(f(a)=c\), write the left and right difference quotients in terms of the branch formulas \(L(x)\) and \(R(x)\).
- Why must a finite derivative at a join imply that the function is continuous there?
- In the second worked example, which parameter condition ensures continuity, and what additional condition ensures differentiability?
- In the third worked example, why do the parameter conditions allow a family rather than a unique choice?
- What error occurs if you compare the slopes of the branch formulas without including \(f(a)\) in the difference quotients?
