Math 323 - Practice Exam III Solutions: Limits, Derivatives, and Contractive Functions | Exams Mathematical Methods for Numerical Analysis and Optimization

November 10, 2001

Math 323 — Practice Exam III

1. (20 points)

(a) Suppose that gis continuous at a t-value a; that there is an interval around a(say Vβ(a)

where β > 0) for which, for all tin the interval except t=a,g(t)6=g(a); and that

limx→g(a)f(x) = L. Prove that limt→a(f◦g)(t) = L.

(b) (This part shows why one of the hypotheses in (a) is needed, and also why we must take

some care in proving the Chain Rule.) Let

f(x) = (0 if x6= 2

1 if x= 2 and g(t) = (tsin 1

t+ 2 if t6= 0

2 if t= 0 .

Prove that limx→2f(x) = 0 and gis continuous at t= 0, but that limt→0(f◦g)(t) doesn’t

exist. (Hint: Consider two sequences of t-values converging to 0, (1/n) and (1/(nπ)).)

2. (20 points) In this problem, you may assume without proof that sin and cos are continuous

functions on all of Rand that limt→0((sin t)/t) = 1.

(a) Use the identity sinA−sin B= 2 sin(1

2(A−B)) cos( 1

2(A+B)) to find the derivative of

sin at the x-value c, from the limit definition of derivative. (In this process, of course,

you are proving that sin is differentiable at c).

(b) Assume that cos is also differentiable, and use the derivative rules and the identity

sin2A+ cos2A= 1 to find the derivative of cos.

3. (30 points) Let Ibe an interval and f:I→Rbe differentiable on all of I. Assume that

f0(x)6= 0 for all x∈I.

(a) Prove that either f0(x)>0 for all x∈Ior f0(x)<0 for all x∈I.

(b) For the remaining parts of this problem, assume f0(x)>0 for all x∈I. Prove that fis

strictly increasing on I.

onto function I→J. Thus, there is an inverse function f−1:J→Idefined by, for

y∈J,f−1(y) is the unique x∈Ifor which f(x) = y. Prove that f−1is also strictly

increasing.

(d) Prove thatf−1is continuous at each y-value d∈J. (So as not to make many cases,

assume dis an interior point of J, so that f−1(d) is an interior point of I.)

Remark: In a way similar to but messier than (d), it is possible to prove that, for each d∈J,

(f−1)0(d) exists and equals 1/f0(f−1(d)).

3.5. (8 points) However we define the function exp(x) on R(which is really exin a different

notation, but we don’t want to assume other properties of that function), we should be able

to verify from that definition that exp(0) = 1 and exp is (differentiable and) equal to its own

derivative for all x∈R. Assume that the inverse function ln of exp is also differentiable, and

use the fact that exp0(x) = exp(x) to find the derivative of ln(y) (with respect to y). (I have

switched to the variable ybecause I think of the domain of exp as a subset of the x-axis and

its range, the domain of ln, as a subset of the y-axis. Hint: What is exp(ln(y))?)

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November 10, 2001 Math 323 — Practice Exam III

(20 points)

(a) Suppose that g is continuous at a t-value a; that there is an interval around a (say Vβ (a) where β > 0) for which, for all t in the interval except t = a, g(t) 6 = g(a); and that limx→g(a) f (x) = L. Prove that limt→a(f ◦ g)(t) = L. (b) (This part shows why one of the hypotheses in (a) is needed, and also why we must take some care in proving the Chain Rule.) Let

f (x) =

{ 0 if x 6 = 2 1 if x = 2 and^ g(t) =

{ t sin (^1) t + 2 if t 6 = 0 2 if t = 0.

Prove that limx→ 2 f (x) = 0 and g is continuous at t = 0, but that limt→ 0 (f ◦g)(t) doesn’t exist. (Hint: Consider two sequences of t-values converging to 0, (1/n) and (1/(nπ)).)

(20 points) In this problem, you may assume without proof that sin and cos are continuous functions on all of R and that limt→ 0 ((sin t)/t) = 1.

(a) Use the identity sin A − sin B = 2 sin( 12 (A − B)) cos( 12 (A + B)) to find the derivative of sin at the x-value c, from the limit definition of derivative. (In this process, of course, you are proving that sin is differentiable at c). (b) Assume that cos is also differentiable, and use the derivative rules and the identity sin^2 A + cos^2 A = 1 to find the derivative of cos.

(30 points) Let I be an interval and f : I → R be differentiable on all of I. Assume that f ′(x) 6 = 0 for all x ∈ I.

(a) Prove that either f ′(x) > 0 for all x ∈ I or f ′(x) < 0 for all x ∈ I. (b) For the remaining parts of this problem, assume f ′(x) > 0 for all x ∈ I. Prove that f is strictly increasing on I. (c) By the Intermediate Value Theorem, f (I) is an interval J. Then by (b), f is a one-to-one onto function I → J. Thus, there is an inverse function f −^1 : J → I defined by, for y ∈ J, f −^1 (y) is the unique x ∈ I for which f (x) = y. Prove that f −^1 is also strictly increasing. (d) Prove thatf −^1 is continuous at each y-value d ∈ J. (So as not to make many cases, assume d is an interior point of J, so that f −^1 (d) is an interior point of I.)

Remark: In a way similar to but messier than (d), it is possible to prove that, for each d ∈ J, (f −^1 )′(d) exists and equals 1/f ′(f −^1 (d)).

3.5. (8 points) However we define the function exp(x) on R (which is really ex^ in a different notation, but we don’t want to assume other properties of that function), we should be able to verify from that definition that exp(0) = 1 and exp is (differentiable and) equal to its own derivative for all x ∈ R. Assume that the inverse function ln of exp is also differentiable, and use the fact that exp′(x) = exp(x) to find the derivative of ln(y) (with respect to y). (I have switched to the variable y because I think of the domain of exp as a subset of the x-axis and its range, the domain of ln, as a subset of the y-axis. Hint: What is exp(ln(y))?)

(22 points) Recall that a function f from a subset A of R into R is called contractive iff there is a constant s ∈ (0, 1) such that, for all x 1 , x 2 ∈ A, we have |f (x 1 ) − f (x 2 )| ≤ s|x 1 − x 2 |.

(a) Prove that a contractive function is uniformly continuous. (b) Prove that, if A is a compact interval [a, b], f ′(x) is continuous on [a, b] and − 1 < f ′(x) < 1 for all x ∈ [a, b], then f is contractive. (c) Give an example to show that, if the interval [a, b] is replaced by the interval [a, ∞) (still closed but no longer bounded) in (b), then the conclusion fails.

(a) Let ε > 0 be given. Then for all x 1 , x 2 ∈ A such that |x 1 − x 2 | < ε, |f (x 1 ) − f (x 2 )| ≤ s|x 1 − x 2 | < |x 1 − x 2 | < ε. (b) Because f ′^ is continuous on the closed interval [a, b], there are x-values c, d ∈ [a, b] for which f ′(c) ≤ f ′(x) ≤ f ′(d) for all x ∈ [a, b], so that |f (x)| ≤ max{|f (c)|, |f (d)|} = s < 1 for all x ∈ [a, b]. Now, for x 1 , x 2 ∈ [a, b], by the MVT there is an x 3 between them for which f (x 1 ) − f (x 2 ) = f ′(x 3 )(x 1 − x 2 ), and so |f (x 1 ) − f (x 2 )| = |f ′(x 3 )||x 1 − x 2 | ≤ s|x 1 − x 2 |. Therefore, f is contractive on [a, b]. (c) Let f be a function whose derivative approaches 1 from below as x → ∞; say f ′(x) = x^2 /(x^2 + 1) = 1 − 1 /(x^2 + 1), so f (x) = x − arctan(x). Then because f ′^ is not bounded away from 1, there is no s < 1 for which |f (x 1 ) − f (x 2 )|/|x 1 − x 2 | ≤ s for all x 1 , x 2 : If we take x 1 = x and x 2 = x − 1, then as x → ∞, |f (x 1 ) − f (x 2 )|/|x 1 − x 2 | = 1 − (arctan(x) − (arctan(x − 1)) approaches 1.

Math 323 - Practice Exam III Solutions: Limits, Derivatives, and Contractive Functions, Exams of Mathematical Methods for Numerical Analysis and Optimization

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