4.4 Inverse function theorem
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Inverse function theorem
Let us start with a simple example. Consider a function f(x):=ax for a numbera6=0. Then f: R→Ris bijective, and the inverse is f−1(y) =1ay. In particular, f0(x) =aand(f−1)0(y) =1a. As
differentiable functions are “infinitesimally like” linear functions, then we expect the same behavior from the inverse function. The main idea of differentiating inverse functions is the following lemma. Lemma 4.4.1. Let I,J ⊂R be intervals. If f: I →J is strictly monotone (hence one-to-one),
onto ( f(I) =J), differentiable at x0∈I, and f0(x0)6=0, then the inverse f−1 is differentiable at y0= f(x0)and
(f−1)0(y0) = f0 f−11(y
If f is continuously differentiable and f0is never zero, then f−1is continuously differentiable. Proof. By , f has a continuous inverse. For convenience call the inverseg: J→I. Letx0,y0be as in the statement. For anyx∈I writey:= f(x). Ifx6=x0and soy6=y0, we find
g(y)−g(y0) y−y0 = g f(x) −g f(x0) f(x)− f(x0) = x−x0 f(x)−f(x0). See for the geometric idea.
x=g(y) x0=g(y0) f(x) =y f(x0) =y0 slope= f(x)−f(x0) x−x0 = y−y0 g(y)−g(y0) f(x) =y f(x0) =y0 x=g(y) x0=g(y0) slope= x−x0 f(x)−f(x0)= g(y)−g(y0) y−y0
Figure 4.10:Interpretation of the derivative of the inverse function.
Let Q(x):= ( x−x0 f(x)−f(x0) ifx6=x0, 1 f0(x0) ifx=x0 (notice that f0(x0)6=0).
As f is differentiable atx0, we have lim x→x0Q(x) =xlim→x0 x−x0 f(x)−f(x0) = 1 f0(x0) =Q(x0),
that is,Qis continuous atx0. Asg(y)is continuous aty0, the compositionQ g(y)
= g(y)−g(y0) y−y0 is continuous aty0by . Therefore 1 f0 g(y0) =Q g(y0) = lim y→y0Q g(y) = lim y→y0 g(y)−g(y0) y−y0 . Sogis differentiable aty0andg0(y0) = f0(g1(y0)).
If f0 is continuous and nonzero at allx∈I, then the lemma applies at allx∈I. Asgis also continuous, the derivativeg0(y) = 1
f0(g(y)) must be continuous.
What is usually called the inverse function theorem is the following result.
Theorem 4.4.2 (Inverse function theorem). Let f: (a,b)→R be a continuously differentiable
function, x0∈(a,b)a point where f0(x0)6=0. Then there exists an open interval I⊂(a,b)with x0∈I, the restriction f|I is injective with a continuously differentiable inverse g: J→I defined on an interval J:= f(I), and
g0(y) = 1
f0 g(y), for all y∈J.
Proof. Without loss of generality, suppose f0(x0)>0. As f0is continuous, there must exist an open intervalI= (x0−δ,x0+δ)such that f0(x)>0 for allx∈I. See .
By f is strictly increasing onI, and hence the restriction f|I bijective onto J:= f(I). As f is continuous, then by the (or directly via the
) f(I)is in interval. Now apply .
If you tried to prove the existence of roots directly as in you saw how difficult that endeavor is. However, with the machinery we have built for inverse functions it becomes an almost trivial exercise, and with the lemma above we prove far more than mere existence.
Corollary 4.4.3. Given any n∈Nand any x ≥0 there exists a unique number y≥0 (denoted x1/n:=y), such that yn=x. Furthermore, the function g: (0,∞)→(0,∞)defined by g(x):=x1/n
is continuously differentiable and
g0(x) = 1 nx(n−1)/n =
nx(1−n)/n, using the convention xm/n:= (x1/n)m.
Proof. Forx=0 the existence of a unique root is trivial.
Let f: (0,∞)→(0,∞)be defined by f(y):=yn. The function f is continuously differentiable and f0(y) =nyn−1, see . For y>0 the derivative f0 is strictly positive and so again by , f is strictly increasing (this can also be proved directly). Given any
M>1, f(M) =Mn≥M, and given any 1>ε >0, f(ε) =εn≤ε. For any xwithε <x<M
we have by the that x∈ f [ε,M]⊂ f (0,∞). AsM and ε were
arbitrary, f is onto (0,∞), and hence f is bijective. Letgbe the inverse of f and we obtain the existence and uniqueness of positiventh roots. saysghas a continuous derivative and g0(x) = 1
Example 4.4.4: The corollary provides a good example of where the inverse function theorem gives us an interval smaller than(a,b). Take f: R→Rdefined by f(x):=x2. Then f0(x)6=0 as long asx6=0. Ifx0>0, we can takeI= (0,∞), but no larger.
Example 4.4.5: Another useful example is f(x):=x3. The function f: R→Ris one-to-one and
onto, so f−1(y) =y1/3exists on the entire real line including zero and negative y. The function f has a continuous derivative, but f−1has no derivative at the origin. The point is that f0(0) =0. See for a graph, notice the vertical tangent on the cube root at the origin. See also
Figure 4.11:Graphs ofx3andx1/3.
Exercise4.4.1: Suppose f:R→Ris continuously differentiable such that f0(x)>0for all x. Show that f
is invertible on the interval J= f(R), the inverse is continuously differentiable, and(f−1)0(y)>0for all
Exercise4.4.2: Suppose I,J are intervals and a monotone onto f:I→J has an inverse g:J→I. Suppose
you already know that both f and g are differentiable everywhere and f0is never zero. Using chain rule but
not prove the formula g0(y) = 1
Exercise4.4.3: Let n∈Nbe even. Prove that every x>0has a unique negative nth root. That is, there
exists a negative number y such that yn=x. Compute the derivative of the function g(x):=y.
Exercise4.4.4: Let n∈Nbe odd and n≥3. Prove that every x has a unique nth root. That is, there exists a
number y such that yn=x. Prove that the function defined by g(x):=y is differentiable except at x=0and
Exercise4.4.5(requires ): Show that if in the inverse function theorem f has k continuous derivatives,
then the inverse function g also has k continuous derivatives.
Exercise4.4.6: Let f(x):=x+2x2sin(1/x)for x6=0and f(0) =0. Show that f is differentiable at all x, that f0(0)>0, but that f is not invertible on any open interval containing the origin.
a) Let f:R→Rbe a continuously differentiable function and k>0be a number such that f0(x)≥k for
all x∈R. Show f is one-to-one and onto, and has a continuously differentiable inverse f−1:R→R.
b) Find an example f:R→Rwhere f0(x)>0for all x, but f is not onto.
Exercise4.4.8: Suppose I,J are intervals and a monotone onto f:I→J has an inverse g:J→I. Suppose x∈I and y:= f(x)∈J, and that g is differentiable at y. Prove:
a) If g0(y)6=0, then f is differentiable at x.