Let f be a twice differentiable function on R with f , f ′ , f ′′ increasing. Fix numbers a and b with a ≤b . For each x >0 , define ξ = ξ (x) so that

5.3.10. Let f be a twice differentiable function on R with f , f ′ , f ′′ increasing. Fix numbers a and b with a ≤b . For each x >0 , define ξ = ξ (x) so that

5.3 The Main Theorems 225

f (b + x) − f (a − x)

=f ′ ( ξ )

b − a + 2x

by the mean value theorem. Prove that ξ (x) is an increasing function of x . Can the hypothesis “ f ′′ increasing” be replaced with “ f ′′ positive”?

M. McAsey and L.A. Rubel, Amer. Math. Monthly, Problem E 3033 Solution. The equation

ξ (x) = ( f ′ ) −1

b − a + 2x

shows that ξ is differentiable. Differentiating the equation defining f ′ ( ξ ) yields

f ′ (b + x) + f ′ (a − x) − 2 f ′ ( ξ ) =f ′′ ( ξ ξ ′

Since f ′ is increasing, f ′′ ≥ 0. Thus ξ ′ (x) > 0 would follow from (5.12), provided that

f ′ (b + x) + f ′ (a − x)

>f ′ ( ξ ).

But f ′′ increasing implies f ′ convex; hence (5.13) holds. However, this is not an obvious step, and we advise the reader to provide details for this statement.

The condition that f ′′ increases cannot be replaced with f ′′ > 0. Indeed, if

f (x) = π x /2 + x arctanx − ln(1 + x 2 )/2, then f and f ′ are increasing, f ′′ > 0, but f ′′ decreases on R + . Setting a = b = 0, we find that the equation defining ξ (x) becomes

π = + arctan ξ (x) ;

hence ξ (x) ≡ 0. ⊓ ⊔ Are there changes if in the Rolle mean value theorem the hypothesis f (a) = f (b)

refers to higher-order derivatives?

be a differentiable function such that f ′ (a) = f ′ (b) . Prove that there exists ξ ∈ (a,b) such that

5.3.11. (a) Let f : [a, b]→R

) − f (a) = ( ′ ξ − a) f ( ξ ).

be a twice differentiable function such that f ′′ (a) = f ′′ (b) . Prove that there exists ξ ∈ (a,b) such that

(b) Let f : [a, b]→R

f ( ξ ) − f (a) = ( ξ − a) f ( ξ − a)

f ′′ ( ξ ).

T. Flett

226 5 Differentiability Solution. (a) Consider the continuous function g : [a, b]→R defined by

g x −a (x) = , if x ∈ (a,b],

f (x)− f (a)

f ′ (a),

if x = a.

If g achieves an extremum at an interior point, say ξ ∈ (a,b), then by Fermat’s theorem, g ′ ( ξ ) = 0 and we conclude the proof. ⊓ ⊔

Let us now assume the contrary, so the only extremum points of g are a and b. Without loss of generality we can suppose that for all x ∈ [a,b] we have g(a) ≤

g (x) ≤ g(b). The second inequality can be rewritten as

f (x) ≤ f (a) + (x − a)g(b), for all x ∈ [a,b].

This yields, for any x ∈ [a,b),

f (b) − f (x)

f (b) − f (a) − (x − a)g(b)

f (b) − f (a)

b −a Taking x ր b we obtain f ′ (b) ≥ g(b). Using the hypothesis we obtain f ′ (a) ≥ g(b),

b −x

b −x

so g (a) ≥ g(b). This implies that g is constant, that is, g ′ = 0 in (a, b). Thus, for all ξ ∈ (a,b), f ( ξ ) − f (a) = ( ξ − a) f ′ ( ξ ).

(b) We have left the details to the reader, since the proof applies the same ideas as above. A key ingredient is a good choice of the auxiliary function g.

A “double” mean value theorem is provided below. The proof combines the stan- dard Lagrange mean value theorem with the basic property that the derivative has the intermediate value theorem (the Darboux theorem).

be real numbers with a <b<c and consider a differentiable function f : [a, c]→R . Prove that there exist ξ ∈ (a,b) , η ∈ (a,c) , ξ < η, such that

5.3.12. Let a , b, c

f (a) − f (b) = (a − b) f ′ ( ξ ) and f (a) − f (c) = (a − c) f ′ ( η ) .