218 CHAPTER 8. IMPLICIT FUNCTION THEOREM

And so

= h′ (t)+m

∑k=1

h(k+1) (t)k!

(1− t)k−m−1

∑k=0

h(k+1) (t)k!

(1− t)k

−K (m+1)(1− t)m

= h′ (t)+h(m+1) (t)

m!(1− t)m−h′ (t)−K (m+1)(1− t)m

and so K = h(m+1)(t)(m+1)! .This proves the theorem. ■

Now let f : U → R where U ⊆ X a normed vector space and suppose f ∈Cm (U) andsuppose Dm+1 f (x) exists on U . Let x ∈U and let r > 0 be such that

B(x,r)⊆U.

Then for ∥v∥< r considerf (x+tv)− f (x)≡ h(t)

for t ∈ [0,1]. Then by the chain rule,

h′ (t) = D f (x+ tv)(v) ; h′′ (t) = D2 f (x+ tv)(v)(v)

and continuing in this way,

h(k) (t) = D(k) f (x+tv)(v)(v) · · ·(v)≡ D(k) f (x+tv)vk.

It follows from Taylor’s formula for a function of one variable given above that

f (x+v) = f (x)+m

∑k=1

D(k) f (x)vk

k!+

D(m+1) f (x+tv)vm+1

(m+1)!. (8.21)

This proves the following theorem.

Theorem 8.6.2 Let f : U → R and let f ∈Cm+1 (U). Then if

B(x,r)⊆U,

and ∥v∥< r, there exists t ∈ (0,1) such that 8.21 holds.

8.7 Second Derivative TestNow consider the case where U ⊆ Rn and f : U → R is C2 (U). Then from Taylor’s theo-rem, if v is small enough, there exists t ∈ (0,1) such that

f (x+v) = f (x)+D f (x)v+D2 f (x+tv)v2

2. (8.22)

Consider

D2 f (x+tv)(ei)(e j) ≡ D(D( f (x+tv))ei)e j

= D(

∂ f (x+ tv)∂xi

)e j =

∂ 2 f (x+ tv)∂x j∂xi