Why Joining Points Need Special Attention
A piecewise-defined function uses different formulas on different parts of its domain. Within each part, finding critical numbers usually means differentiating the applicable formula and solving for where its derivative is zero. But a change of formula creates another possibility: the derivative may fail to exist at the joining point itself.
To find all critical numbers, therefore, we need to check two kinds of locations: points inside the intervals where each formula applies, and joining points that belong to the domain. A join can be smooth, have a corner, or be discontinuous. These cases have different effects on differentiability.
Critical Numbers and Piecewise Definitions
A number \(c\) is a critical number of a function \(f\) if \(c\) is in the domain of \(f\) and either \(f'(c)=0\) or \(f'(c)\) does not exist. The requirement that \(c\) be in the domain matters: a point where the function is undefined cannot be a critical number.
For functions whose domain includes endpoints, the ordinary derivative is a two-sided limit, so it may not exist at an endpoint. Such endpoints are often listed separately when considering candidates for extrema. Here, the main concern is a joining point inside the domain, where values from both sides can be checked.
Suppose a formula applies on an interval \(I\). Its derivative can produce candidate critical numbers only in \(I\), and only where the function is defined. A solution of \(f'(x)=0\) outside the interval on which that formula applies is not a candidate from that branch. After checking each interval, test every joining point that belongs to the domain: evaluate the function there, compare its one-sided limits, and, if it is continuous, compare the one-sided derivatives.
A Test for Differentiability at a Join
At an interior joining point \(c\), the derivative is defined by the two-sided difference quotient. The left-hand derivative uses values with \(x<c\), and the right-hand derivative uses values with \(x>c\). If these one-sided derivatives both exist as finite numbers, the ordinary derivative exists precisely when they are equal.
Continuity must also be checked. Differentiability at \(c\) implies continuity there: if \(f'(c)\) exists, then
As \(x\) approaches \(c\), the quotient on the right approaches the finite number \(f'(c)\), while \(x-c\) approaches zero. Their product therefore approaches zero, so \(f(x)\) approaches \(f(c)\). Thus, if a function is discontinuous at an included join, it cannot be differentiable there.
Conversely, continuity alone does not guarantee differentiability. A continuous join can still have different one-sided slopes. The following examples show how to apply both checks.
Worked Examples
Worked Example: A Smooth Joining Point
Find the critical numbers of
The first formula applies for \(x<1\), and the second for \(x>1\). Differentiate within those intervals:
On the first interval, \(2x=0\) gives \(x=0\), and \(0<1\), so it is a valid candidate. On the second interval, the derivative is \(2\), which is never zero. Thus there are no other interior candidates from the branches.
Now check the join \(x=1\). Its function value comes from the first branch:
The left-hand limit is \(1\), and the right-hand limit is \(2(1)-1=1\), so the function is continuous at the join. The one-sided derivatives are
They agree, so \(f'(1)=2\). The join is not a critical number. The only critical number is \(x=0\). Check: \(f(0)=0\), and in a neighborhood of \(0\), \(f(x)=x^2\), whose derivative is \(2x\); hence \(f'(0)=0\), as required.
Worked Example: A Continuous Join with Different Slopes
Find the critical numbers of
For \(x<0\), \(g'(x)=2x\). Its only zero is \(x=0\), but that point is not in this branch’s interval. For \(x>0\), \(g'(x)=1\), which has no zeros. Neither branch therefore supplies an interior critical number.
The join \(x=0\) is in the domain, and \(g(0)=0\). The left-hand limit is \(0^2=0\), while the right-hand limit is \(0\), so \(g\) is continuous there. The one-sided derivatives are
Because \(0\ne 1\), the derivative at the join does not exist. Since \(0\) belongs to the domain, it is a critical number. Check: the left and right difference quotients approach different values, so there is no two-sided derivative at \(0\). Thus the only critical number is \(x=0\).
Worked Example: A Discontinuous Joining Point
Find the critical numbers of
On \(x<1\), the derivative is \(1\), so it has no zeros there. On \(x>1\), the derivative is also \(1\), so this branch has no interior zeros either.
At \(x=1\), the function is defined by the second branch, so \(h(1)=2\). The left-hand limit is \(1\), but the right-hand limit is \(2\). The function is discontinuous at \(1\). Since differentiability implies continuity, \(h'(1)\) does not exist. The point \(1\) is in the domain, so it is a critical number. Check: neither branch has a zero derivative, and the only join is \(1\); consequently, the complete list is \(x=1\).
A Reliable Procedure
For a piecewise-defined function with finitely many formulas, use the following process. It separates ordinary branch calculations from the additional analysis required at joins.
Note which formula applies on each interval and whether each joining point belongs to the domain.
Use the derivative of that interval’s formula. Solve for zeros and keep only solutions lying in the interval where the formula applies.
Compare the one-sided limits with the function’s value at the join. A discontinuous included join is a point where the derivative does not exist.
If both are finite and equal, the derivative exists there. If they differ, the derivative does not exist, so the join is a critical number.
Include only points in the domain where the derivative is zero or does not exist. Check that no branch root was retained outside its valid interval.
Common Pitfall: Checking Slopes Without Checking Values
A tempting shortcut is to compare the slopes of the two formulas and treat that comparison as the entire join test. It is not enough. In the discontinuous example, both branches have slope \(1\), but the function jumps at \(x=1\). Equal slopes of the branch formulas do not establish that the function is differentiable at the join.
A related mistake is to solve a branch’s derivative equation and keep every numerical solution. The branch formula is valid only on its stated interval. In the continuous-corner example, the equation \(2x=0\) gives \(x=0\) for the left branch, but that branch applies only when \(x<0\). The point is not a critical number because of that equation; it is critical because the derivative fails to exist at the join.
Why This Analysis Matters
A piecewise function can change its local behavior not only where a branch derivative becomes zero, but also where the formula changes. A smooth join contributes no critical number unless its derivative is zero. A corner or a jump can contribute one because the derivative does not exist, provided the joining point belongs to the domain.
This is why checking only the derivative formulas on open intervals can miss critical numbers. The complete analysis accounts for every interval and every included join, while respecting the domain and the exact condition in the definition.
Check Your Understanding
For each question, explain which branch or joining-point check is needed.
- Why must a solution of a branch’s derivative equation lie in the interval where that branch applies?
- What must be checked at a joining point before comparing one-sided derivatives?
- If a function is continuous at a join but the one-sided derivatives are unequal, what follows about the derivative there?
- Can an included discontinuous joining point be a critical number? Explain why.
- In the function \(q(x)=x^2\) for \(x<2\) and \(q(x)=3x-2\) for \(x\geq 2\), what two checks determine whether \(x=2\) is critical?
