AP CALCULUS ABDifferentiability of Piecewise Defined Functions

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Authored by

  • Jim Hartman
    College of Wooster
    Wooster, Ohio
  • Larry Riddle
    Agnes Scott College
    Decatur, Georgia

Theorem 1: Suppose g is differentiable on an open interval containing x=c. If both stack l i m space g apostrophe left parenthesis x right parenthesis with x rightwards arrow c minus below and stack l i m space g apostrophe left parenthesis x right parenthesis with x rightwards arrow c plus below exist, then the two limits are equal, and the common value is g'(c).

Proof: Let stack L subscript 1 equals l i m space g apostrophe left parenthesis x right parenthesis with x rightwards arrow c minus below and stack L subscript 2 equals l i m space g apostrophe left parenthesis x right parenthesis with x rightwards arrow c plus below. By the Mean Value Theorem, for every positive h sufficiently small, there exists d subscript h satisfying c less than d subscript h less than c plus h such that: fraction numerator g left parenthesis c plus h right parenthesis minus g left parenthesis c right parenthesis over denominator h end fraction equals g apostrophe left parenthesis d subscript h right parenthesis.

Then: g apostrophe left parenthesis c right parenthesis equals stack l i m with h rightwards arrow 0 plus below fraction numerator g left parenthesis c plus h right parenthesis minus g left parenthesis c right parenthesis over denominator h end fraction equals stack l i m with h rightwards arrow 0 plus below g apostrophe left parenthesis d subscript h right parenthesis equals stack l i m with h rightwards arrow c plus below g apostrophe left parenthesis x right parenthesis equals L subscript 2.

Similarly, for every positive h sufficiently small, there exists e subscript h satisfying c minus h less than e subscript h less than c such that: fraction numerator g left parenthesis c right parenthesis minus g left parenthesis c minus h right parenthesis over denominator h end fraction equals g apostrophe left parenthesis e subscript h right parenthesis.

Then: g apostrophe left parenthesis c right parenthesis equals stack l i m with h rightwards arrow 0 plus below fraction numerator g left parenthesis c right parenthesis minus g left parenthesis c minus h right parenthesis over denominator h end fraction equals stack l i m space g apostrophe left parenthesis e subscript h right parenthesis with h rightwards arrow 0 plus below equals stack l i m space g apostrophe left parenthesis x right parenthesis equals L subscript 1 with x rightwards arrow c minus below. This shows that L subscript 1 equals L subscript 2 equals g apostrophe left parenthesis c right parenthesis.

Note: The same proof can be modified to show that if g is continuous at x=c and differentiable on both sides of x=c, and if stack l i m space g apostrophe left parenthesis x right parenthesis with x rightwards arrow c minus below equals stack l i m space g apostrophe left parenthesis x right parenthesis equals L with x rightwards arrow c plus below, then g is differentiable at x=c with g'(c)=L.

Theorem 2: Suppose p and q are defined on an open interval containing x=c, and each are differentiable at x=c.

Let: f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell p left parenthesis x right parenthesis space f o r space a less than x less or equal than c end cell row cell q left parenthesis x right parenthesis space f o r space c less than x less than b end cell end table close

Then f is differentiable at x=c if and only if p(c)=q(c) and p'(c)=q'(c).

Proof: We know that f'(c) exists if and only if stack l i m with x rightwards arrow c minus below fraction numerator f left parenthesis x right parenthesis minus f left parenthesis c right parenthesis over denominator x minus c end fraction equals stack l i m with x rightwards arrow c plus below fraction numerator f left parenthesis x right parenthesis minus f left parenthesis c right parenthesis over denominator x minus c end fraction.

We have that: stack l i m with x rightwards arrow c minus below fraction numerator f left parenthesis x right parenthesis minus f left parenthesis c right parenthesis over denominator x minus c end fraction equals stack l i m with x rightwards arrow c minus below fraction numerator p left parenthesis x right parenthesis minus p left parenthesis c right parenthesis over denominator x minus c end fraction equals p apostrophe left parenthesis c right parenthesis.

Also: stack l i m with x rightwards arrow c plus below fraction numerator f left parenthesis x right parenthesis minus f left parenthesis c right parenthesis over denominator x minus c end fraction equals stack l i m with x rightwards arrow c plus below fraction numerator q left parenthesis x right parenthesis minus p left parenthesis c right parenthesis over denominator x minus c end fraction equals stack l i m with x rightwards arrow c plus below open parentheses fraction numerator q left parenthesis x right parenthesis minus q left parenthesis c right parenthesis over denominator x minus c end fraction plus fraction numerator q left parenthesis x right parenthesis minus p left parenthesis c right parenthesis over denominator x minus c end fraction close parentheses equals q apostrophe left parenthesis c right parenthesis if and only if p(c)=q(c). So f will be differentiable at x=c if and only if p(c)=q(c) and p'(c)=q'(c).

2003 AB6, part (c)

Suppose the function g is defined by: g left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell k square root of x plus 1 end root space f o r space 0 less or equal than x less or equal than 3 end cell row cell m x plus 2 space f o r space 3 less than x less or equal than 5 end cell end table close where k and m are constants. If g is differentiable at x=3 what are the values of k and m?

Method 1:

We are told that g is differentiable at x=3, and so g is certainly differentiable on the open interval (0,5).

stack l i m space g apostrophe left parenthesis x right parenthesis with x rightwards arrow 3 to the power of minus below equals stack l i m space with x rightwards arrow 3 to the power of minus below fraction numerator k over denominator 2 square root of x plus 1 end root end fraction equals k over 4 and stack l i m space g apostrophe left parenthesis x right parenthesis with x rightwards arrow 3 to the power of plus below equals stack l i m space m with x rightwards arrow 3 to the power of plus below equals m.

So the two limits both exist and by Theorem 1 must be equal. Hence k over 4 equals m. Since g is continuous at x=3, 2k=3m+2. This gives the two equations to solve for k and m.

Method 2:

Let p left parenthesis x right parenthesis equals k square root of x plus 1 end root and q(x)=mx+2. Both are differentiable at x=3. If g is differentiable at x=3, then Theorem 2 implies that p(3)=q(3) and p'(3)=q'(3). This yields the two same two equations as Method 1.

Either the note after Theorem 1 or Theorem 2 can be used to show that if we choose k equals 8 over 5 and m equals 2 over 5, then we can prove that g is differentiable at x=3.

Jim Hartman teaches at the College of Wooster in Wooster, Ohio.

Larry Riddle is a professor of mathematics at Agnes Scott College in Decatur, Georgia.