McGill University Department of Mathematics and Statistics
MATH 243 Analysis 2, Winter 2023
Assignment 2: Solutions
1. Use the intermediate value theorem and Rolle’s theorem to prove that the function f : R+ → R, f(x) := x3 +2×2 −5+ x1 has exactly 2 roots in R+. You are not permitted to use any Calculator- based arithmetic in your solution.
Note that f is differentiable on R+, and thus differentiable and continuous on any closed and bounded subinterval of R+. We can thus freely apply the intermediate value theorem and Rolle’s theorem to any such subinterval.
We will first prove that f has at least two roots in R+. Observe that 1 13 12
f(5)= 5 +2 5 −5+5>0 f(1) = 1 + 2 − 5 + 1 = −1 < 0
f(2) = 8 + 8 − 5 + 12 > 0
Hence f has opposite signs at the endpoints of [ 15 , 1] and at the endpoints of [1, 2]. It now follows from the intermediate value theorem that f has at least one root in ]15,1[ and at least one root in ]1,2[, which proves that f has at least two roots in R+.
We will now prove that f has at most two roots in R+. We will use a proof by contradiction:
assume that f has at least three roots in R+, say at x1, x2, x3, where 0 < x1 < x2 < x3. Applying
Rolle’stheoremtofontheintervals[x,x]and[x,x]yieldsthatf′(x)=3x2+4x− 1 has 12 23 x2
at least one root in ]x1,x2[ and at least one root in ]x2,x3[. This means that f′ has at least two rootsinR+,sayatt1 andt2,where0
Hint: Prove that for any δ > 0 there exists an x ∈]−δ,δ[, x ̸= 0, such that f′(x) < 0. Then, using the fact that f′ is continuous at x, prove that there exists an η > 0 such that f is decreasing on ]x−η,x+η[⊆]−δ,δ[.
(a) Let x ̸= 0. Then
f ′ (x) = 1 + 4x sin(1/x) + 2×2 cos(1/x) · (−1/x2 ) = 1 + 4x sin(1/x) − 2 cos(1/x)
Since all functions involved are continuous on R \ {0}, f is continuously differentiable on R\{0}. Atx=0wehave
f′(0)= lim f(x)−f(0) = lim x+2x2sin(1/x) = lim(1+2xsin(1/x)) x→0 x−0 x→0 x x→0
= 1 + 2 lim (x sin(1/x)) = 1 + 2 · 0 = 1 x→0
since lim x = 0 and sin(1/x) is bounded on R \ {0} (note that | sin(1/x)| ≤ 1 for all x→0
x ∈ R \ {0}). Thus f is differentiable at 0. Summarizing, f is continuously differentiable on R \ {0} and differentiable at 0 with f′(0) = 1.
Remark: f′ is indeed discontinuous at x = 0. This can be seen e.g. in the following two ways:
(i) Assume that f′ is continuous at 0. Then
1=f′(0)= limf′(x)= lim1+4xsin(1/x)−2cos(1/x)=1+0−2limcos(1/x)
x→0 x→0 x→0
which implies that lim cos(1/x) = 0 but we know that this limit does not exit. Hence
f′ is discontinuous at x = 0.
(ii) Assume that f′ is continuous at 0. Then f′ is continuous on R with f′(0) > 0. It was shown in class that there exists a δ > 0 such that f is strictly increasing on ] − δ, δ[. But it will be shown in part (b) that no such δ exists. Hence f′ is discontinuous at x = 0.
(b) Consider the sequence (xn) where xn := 1 . Then lim (xn) = 0 and 2πn
f′(xn) = 1 + 4xn sin(2πn) − 2 cos(2πn) = 1 + 0 − 2 = −1
for all n ∈ N. For each δ > 0 there thus exists an x ∈ ] − δ, δ[ \{0} (actually infinitely many such x) with f′(x) < 0. It has been shown in class that this implies the existence of an η > 0 such that f is strictly decreasing on ]x − η, x + η[ ⊆ ] − δ, δ[. Hence f is not increasing on any neighborhood of 0.
Here is a graph of the function f, together with the graph of y = x: 2
0 if x = 0
Code Help
-0.10 -0.05
Programming Help, Add QQ: 749389476
3. Consider the function f : R → R, defined by
x4 sin(1/x) if x ̸= 0
(a) Show that f is continuously differentiable.
(b) Let g(x) := 2×4 + f (x). Show that g has an absolute minimum at 0 but that, nonetheless, there does not exist any δ > 0 such that g is decreasing on ] − δ, 0[ and increasing on ]0, δ[.
0 if x = 0
(a) Let x ̸= 0. Then
f′(x) = 4×3 sin(1/x) + x4 cos(1/x) · (− 1 ) = 4×3 sin(1/x) − x2 cos(1/x)
Since all functions involved are continuous on R \ {0}, f is continuously differentiable on
R\{0}. Atx=0wehave
f′(0) = lim f(x) − f(0) = lim x4 sin(1/x) = lim x3 sin(1/x) = 0
x→0 x−0 x→0 x x→0
since lim x3 = 0 and sin(1/x) is bounded on R \ {0} (note that | sin(1/x)| ≤ 1 for all
x ∈ R \ {0}). Thus f is differentiable at 0. It remains to be shown that f ′ is continuous at
lim f′(x) = lim (4×3 sin(1/x) − x2 cos(1/x)) = 4 lim(x3 sin(1/x)) − lim(x2 cos(1/x)) = 0
x→0 x→0 x→0 x→0
(again, since sin and cos are bounded). Thus f is continuously differentiable on R.
(b) We show first that g has an absolute minimum at 0. Let x ̸= 0; then g(x)=2×4 +x4sin(1/x)=x4(2+sin(1/x))≥x4(2−1)=x4 >0
Atx=0wehaveg(0)=0. Thusghasanabsoluteminimumat0(anditistheonly absolute minimum g has).
Now we’ll analyze the intervals of increase and decrease of g: g′(x) = 8×3 + f′(x). Thus
Note that by part (a), g′ is continuous on R \ {0}.
x2(8x + 4x sin(1/x) − cos(1/x)) if x ̸= 0 0 if x = 0
Consider the sequence (xn), where xn := 1 . Then lim (xn) = 0 and 2nπ
′ 184 14
g (xn) = (2nπ)2 2nπ + 2nπ sin(2nπ) − cos(2nπ) = (2nπ)2 nπ − 1 < 0
for all n > 1. For any δ > 0 there thus exists an (even infinitely many) x ∈]0,δ[ with g′(x) < 0 which implies (since g′ is continuous at x) that g is decreasing in a neighborhood of this x. Hence there is no δ > 0 such that g is increasing on ]0, δ[.
Now consider the sequence (xn) where xn := − 1 . Then lim (xn) = 0 and (2n + 1)π
′184 g(xn)=((2n+1)π)2 −(2n+1)π−(2n+1)πsin((2n+1)π)−cos((2n+1)π)
= ((2n+1)π)2 −(2n+1)π +1 >0
for all n. For any δ > 0 there thus exists an (even infinitely many) x ∈]−δ,0[ with g′(x) > 0 which implies (since g′ is continuous at x) that g is increasing in a neighborhood of this x. Hence there is no δ > 0 such that g is decreasing on ] − δ, 0[.
Here is a graph of the function g, oscillating between the graphs of x4 and 3×4: 2.5 × 10-6
2. × 10-6 1.5 × 10-6 1. × 10-6 5. × 10-7
-0.03 -0.02 -0.01
0.01 0.02 0.03
浙大学霸代写 加微信 cstutorcs
4.LetIbeanintervalandletf:I→RbedifferentiableonI. Provethatiff′(x)̸=0forall x∈I,theneitherf′(x)>0forallx∈I orf′(x)<0forallx∈I.
Assume that the statement is wrong. Then there exist a,b ∈ I with f′(a) < 0 < f′(b). Assume w.l.o.g. that a < b; applying Darboux’s theorem to f on the interval [a,b] yields that there exists a c ∈]a,b[ with f′(c) = 0 which is a contradiction. Thus we either have that f′(x) > 0 for allx∈I orthatf′(x)<0forallx∈I.
5. Letf:[0,2]→Rbedifferentiablewithf(0)=0,f(1)=2andf(2)=1. (a) Prove that there exists a c1 ∈]0,2[ with f′(c1) = 21.
(b) Prove that there exists a c2 ∈]0,2[ with f′(c2) = −21. Solution:
(a) Applying the mean value theorem to f on the interval [0, 2] yields that there exists a c1 ∈ ]0, 2[ with f′(c1) = f(2) − f(0) = 1.
(b) We first apply the mean value theorem to f on the intervals [0, 1] and [1, 2].
Applying the mean value theorem to f on the interval [0,1] yields that there exists an a ∈ ]0, 1[ with f ′ (a) = f (1) − f (0) = 2.
And applying the mean value theorem to f on the interval [1, 2] yields that there exists a
b ∈ ]1, 2[ with f ′ (b) = f (2) − f (1) = −1. 2−1
Now we apply Darboux’s theorem to f on the interval [a, b] (note that a < b by construction): we have that f′(a) = 2 > −12 > −1 = f′(b). Thus there exists a c2 ∈]a,b[⊂]0,2[ with f ′ ( c 2 ) = − 21 .