Almaz Khusnutdinov•Math enthusiast, love coding and mathematics
Published 10/20/2025
Rolles Theorem
Problem 1.1
Rolle’s Theorem. Let f:[a,b]→R be continuous on [a,b] and differentiable on (a,b),f(a)=f(b).Prove that ∃c∈(a,b) with f′(c)=0.Outline and detailed explanation: Continuity on a compact interval gives extrema: pick xmin,xmax∈[a,b] with f(xmin)≤f(x)≤f(xmax).If f is constant, then f′(x)=0∀x∈(a,b),so pick any c.Else f(xmin)<f(xmax)or vice versa. Since f(a)=f(b),some strict extremum lies in (a,b).At an interior extremum and differentiability, the derivative vanishes: f′(c)=0.Answer: ∃c∈(a,b) with f′(c)=0.
Problem 1.2
Let f(x)=x3−3xon [−3,3].Verify Rolle’s hypotheses and find all c∈(−3,3)with f′(c)=0.Answer: f(−3)=f(3)=0,f∈C1.f′(x)=3x2−3=3(x2−1)⇒c=±1.
Problem 1.3
Let f(x)=sin(2x)on [0,π].Check f(0)=f(π)and find all c∈(0,π)with f′(c)=0.Answer: f(0)=f(π)=0.f′(x)=2cos(2x)=0⇒2x=2π,23π⇒c=4π,43π.
Problem 1.4
Let f(x)=cosxon [0,2π].Verify Rolle’s conditions and find c∈(0,2π)with f′(c)=0.Answer: f(0)=f(2π)=1.f′(x)=−sinx=0⇒x=kπ.Interior c=π.
Problem 1.5
Let f(x)=x4−2x2on [−1,1].Verify Rolle’s conditions and find all c∈(−1,1)with f′(c)=0.Answer: f(−1)=f(1)=−1.f′(x)=4x3−4x=4x(x2−1).Interior zero c=0.
Problem 1.6
Let f(x)=ln(1+x)−(ln2)xon [0,1].Verify f(0)=f(1)=0and find c∈(0,1)with f′(c)=0.Answer: f′(x)=1+x1−ln2,f′(c)=0⇒1+c1=ln2⇒c=ln21−1∈(0,1).
Problem 1.7
Mean Value Theorem from Rolle. Let f:[a,b]→R be continuous on [a,b] and differentiable on (a,b).Prove that ∃c∈(a,b) with f′(c)=b−af(b)−f(a).Answer (outline): Define the chord ℓ(x)=f(a)+b−af(b)−f(a)(x−a).Set F(x)=f(x)−ℓ(x).Then F(a)=F(b)=0,and F is C1.By Rolle, ∃c:F′(c)=0.F′(c)=f′(c)−b−af(b)−f(a)=0⇒f′(c)=b−af(b)−f(a).
Problem 1.8
Cauchy Mean Value Theorem. Let f,g:[a,b]→R be continuous on [a,b] and differentiable on (a,b).Prove that ∃c∈(a,b) with (f(b)−f(a))g′(c)=(g(b)−g(a))f′(c).Answer (outline): Define H(x)=(f(b)−f(a))g(x)−(g(b)−g(a))f(x).H(a)=H(b).By Rolle, ∃c:H′(c)=0.Compute H′(c)=(f(b)−f(a))g′(c)−(g(b)−g(a))f′(c)=0.Rearrange to the claimed identity.
Problem 1.9
Monotonicity criterion. Let f:[a,b]→R be continuous on [a,b] and differentiable on (a,b).Assume f′(x)≥0∀x∈(a,b).Prove that f is nondecreasing on [a,b].Answer (outline): For any x<y in [a,b],apply the MVT to f on [x,y]:∃c∈(x,y):f(y)−f(x)=f′(c)(y−x).With f′(c)≥0 and y−x>0,⇒f(y)−f(x)≥0.Hence f(x)≤f(y).
Problem 1.10
(a) Tangent parallel to any secant. Let f:[a,b]→R be continuous on [a,b] and differentiable on (a,b).Fix x1<x2 in [a,b].Prove there exists c∈(x1,x2) with f′(c)=x2−x1f(x2)−f(x1).Answer: Use the same chord-subtraction method as in the MVT proof.(b) Prescribed slope from two intersections. Let f:[a,b]→R be continuous on [a,b] and differentiable on (a,b).Let m∈R and suppose x1<x2 in [a,b] satisfy f(x1)−mx1=f(x2)−mx2.Prove there exists c∈(x1,x2) with f′(c)=m.Answer: Set F(x)=f(x)−mx.Then F(x1)=F(x2)=0.By Rolle there exists c with F′(c)=0,so f′(c)=m.(c) Symmetry of derivatives about the midpoint. Let f:[a,b]→R be continuous on [a,b] and differentiable on (a,b),assume f(a)=f(b).Prove there exists c∈(a,b) with f′(c)=f′(a+b−c).Answer: Set G(x)=f(x)−f(a+b−x).Then G(a)=G(b)=0.By Rolle there exists c with G′(c)=0,so f′(c)=f′(a+b−c).
Problem 1.11
Let f(x)=x5−5x3+4xon [−1,1].(a) Verify the hypotheses of Rolle’s theorem for f on [−1,1].(b) Find all c∈(−1,1) such that f′(c)=0.Detailed solution:(i) f is a polynomial, hence continuous on [−1,1] and differentiable on (−1,1).(ii) Endpoint values: f(1)=1−5+4=0,f(−1)=−1+5−4=0,so f(1)=f(−1).Therefore Rolle’s theorem applies and there exists at least one c∈(−1,1) with f′(c)=0.(iii) Compute f′(x)=5x4−15x2+4. Put y=x2(y∈[0,1]),then 5y2−15y+4=0.Solve the quadratic: y=1015±145.Numerically 1015+145≈2.704>1(discard),and 1015−145≈0.29584∈(0,1).Thus x=±1015−145are the only solutions in (−1,1).Answer: c=±1015−145(≈±0.543).
Problem 1.12
Define G(x)=∫0xet2dt−x∫01et2dton [0,1].(a) Show G(0)=G(1).(b) Use Rolle’s theorem to deduce the existence of c∈(0,1) satisfying an equation for c.Answer: G(0)=G(1)=0,and there exists a unique c∈(0,1) with ec2=∫01et2dt.
Problem 1.13
Let f be twice differentiable on [a,b], and suppose there exists m∈(a,b) with f(a)=f(m)=f(b).Show there exists c∈(a,b) with f′′(c)=0.Answer (proof outline):Apply Rolle’s theorem to f on [a,m] to get c1∈(a,m) with f′(c1)=0.Apply Rolle’s theorem to f on [m,b] to get c2∈(m,b) with f′(c2)=0.Now f′ is continuous on [c1,c2] and differentiable on (c1,c2),with f′(c1)=f′(c2)=0.Apply Rolle’s theorem to f′ on [c1,c2] to obtain c∈(c1,c2)⊂(a,b) with f′′(c)=0.
Problem 1.14
Let f(x)=∣x2−1∣ on [−2,2].(a) Are the hypotheses of Rolle’s theorem satisfied on [−2,2]?(b) Find all c∈(−2,2) with f′(c)=0 (when the derivative exists).Answer: f(−2)=f(2)=3 and f is continuous on [−2,2], but f is not differentiable at x=±1(where x2−1=0),so Rolle’s theorem does not strictly apply. Nevertheless, on (−1,1) we have f(x)=1−x2,f′(x)=−2x,hence f′(0)=0 and c=0 is an interior point with f′(c)=0.
Problem 1.15
Let f(x)=∣x∣+∣x−1∣ on [0,1].(a) Verify the hypotheses of Rolle’s theorem.(b) Find all c∈(0,1) such that f′(c)=0.Answer: f(0)=1=f(1),f is continuous on [0,1] and differentiable on (0,1).For x∈(0,1) we have f(x)=x+(1−x)=1⇒f′(x)=0for every x∈(0,1).Hence every c∈(0,1) satisfies f′(c)=0.
Problem 1.16
Let f be continuous on [a,b] and differentiable on (a,b).Assume f has n(≥2) distinct zeros a≤x1<x2<⋯<xn≤b.Prove that f′ has at least n−1 distinct zeros in (a,b).Detailed proof:For each k=1,2,…,n−1 consider the closed interval [xk,xk+1].On [xk,xk+1] the function f is continuous and differentiable on (xk,xk+1), and f(xk)=0=f(xk+1).By Rolle’s theorem there exists ck∈(xk,xk+1) such that f′(ck)=0.The open intervals (xk,xk+1) are pairwise disjoint, so the points c1,c2,…,cn−1 are distinct.Therefore f′ has at least n−1 distinct zeros in (a,b),as required.
Problem 1.17
Let φ be continuous on [0,1] and suppose ∫01φ(t)dt=0.Define F(x)=∫0xφ(t)dt(x∈[0,1]).Use Rolle’s theorem on F.Answer: ∃c∈(0,1) such that φ(c)=0.
Problem 1.18
Let f be continuous on [0,2] and suppose ∫01f(t)dt=∫12f(t)dt.Define G(x)=∫xx+1f(t)dtfor x∈[0,1].Use Rolle’s theorem on G.Answer: ∃c∈(0,1) such that f(c)=f(c+1).
Problem 1.19
Let p(x)=ax3+bx2+cx+d be a cubic polynomial satisfying p(0)=p(1)=p(2).Show that p′′ vanishes at a point in (0,2) and find that point explicitly.Answer: p′′(x)=6ax+2b and the constraints p(0)=p(1)=p(2) force b=−3a,so p′′(x)=6a(x−1).Hence p′′(1)=0(the unique root lies at x=1∈(0,2)).
Mean Value Theorem
Problem 2.1
Let f(x)=x3−3x+1 on [0,2].(a) Verify the hypotheses of the Mean Value Theorem on [0,2].(b) Find all c∈(0,2) such that f′(c)=2f(2)−f(0).Detailed solution:(a) f is a polynomial, hence continuous on [0,2] and differentiable on (0,2). Thus MVT hypotheses hold.(b) Compute f(2)=23−3⋅2+1=8−6+1=3,f(0)=0−0+1=1.Average slope =2−0f(2)−f(0)=23−1=1.f′(x)=3x2−3. We solve 3x2−3=1⟹3x2=4⟹x2=34.Thus x=±32. Only x=32≈1.1547 lies in (0,2).Answer: c=32.
Problem 2.2
Let f(x)=lnx on [1,e].Show MVT applies and find c∈(1,e) with f′(c)=e−1f(e)−f(1).Answer: f′(x)=x1,e−1f(e)−f(1)=e−11−0=e−11,so c=e−1.
Problem 2.3
State and give a proof outline of Cauchy’s Mean Value Theorem:If f,g are continuous on [a,b] and differentiable on (a,b) and g(b)=g(a),then there exists c∈(a,b) such that g(b)−g(a)f(b)−f(a)=g′(c)f′(c).Outline of proof:Define h(x):=f(x)−λg(x) where λ=g(b)−g(a)f(b)−f(a).Then h(a)=h(b). Apply Rolle’s theorem to h to obtain h′(c)=0 for some c∈(a,b).Hence f′(c)−λg′(c)=0⇒g′(c)f′(c)=λ=g(b)−g(a)f(b)−f(a).
Problem 2.4
Let f∈C2[a,b] with ∣f′′(x)∣≤M for all x∈[a,b].Show that ∣f′(b)−f′(a)∣≤M(b−a).Answer (outline): Apply the Mean Value Theorem to f′ on [a,b].There exists c∈(a,b) such that f′(b)−f′(a)=f′′(c)(b−a).Hence ∣f′(b)−f′(a)∣=∣f′′(c)∣(b−a)≤M(b−a).
Problem 2.5
Let f(x)=x1/3 on [−1,1].(a) Compute the average slope S=1−(−1)f(1)−f(−1).(b) Determine whether there exists c∈(−1,1) with f′(c)=S, and explain whether MVT applies.Answer: f(1)=1,f(−1)=−1⇒S=21−(−1)=1.For x=0,f′(x)=31x−2/3.Solve 31x−2/3=1⇒x=3−3/2.Since 3−3/2∈(0,1), such a c exists (so the conclusion holds), but f′ is not defined at 0,so the hypotheses of the MVT fail on (−1,1) (MVT does not apply globally), even though the conclusion happens to hold here.
Problem 2.6
Let f(x)=x4−4x3+6x2 on [0,2].(a) Verify the hypotheses of the Mean Value Theorem on [0,2].(b) Find all c∈(0,2) such that f′(c)=2−0f(2)−f(0).Detailed solution:(a) f is a polynomial, hence continuous on [0,2] and differentiable on (0,2). Thus MVT applies.(b) Compute f(2)=24−4⋅23+6⋅22=16−32+24=8,f(0)=0.Average slope =2−0f(2)−f(0)=28=4.f′(x)=4x3−12x2+12x=4x(x2−3x+3).Solve 4x(x2−3x+3)=4⟹x(x2−3x+3)=1.Rearrange: x3−3x2+3x−1=0⟹(x−1)3=0.Thus the only solution is x=1∈(0,2).Answer: c=1.
Problem 2.7
Let f(x)=x+x2 on [1,4].(a) Verify MVT hypotheses.(b) Find c∈(1,4) with f′(c)=4−1f(4)−f(1).Answer: f′(x)=1−x22,f(4)=4+21=29,f(1)=3.Average slope =329−3=323=21.Solve 1−x22=21⇒x22=21⇒x2=4⇒x=2.Answer: c=2.
Problem 2.8
Let f(x)=sin(2x) on [6π,2π].(a) Check MVT hypotheses.(b) Find the unique c∈(6π,2π) with f′(c)=π/2−π/6f(π/2)−f(π/6).Answer: f′(x)=2cos(2x),f(2π)=0,f(6π)=23.Average slope =2π−6π0−23=3π−23=−2π33.Solve 2cos(2c)=−2π33⇒cos(2c)=−4π33.Answer: c=21arccos(−4π33)(this value lies in (6π,2π)).
Problem 2.9
Let f(x)=ln(1+x2) on [−1,2].(a) Verify MVT hypotheses.(b) Find c∈(−1,2) such that f′(c)=2−(−1)f(2)−f(−1).Answer: f′(x)=1+x22x,f(2)=ln5,f(−1)=ln2.Average slope k=3ln5−ln2≈0.3054302.Solve 1+c22c=k⇒kc2−2c+k=0.Hence c=k1±1−k2.Only c=k1−1−k2≈0.1565 lies in (−1,2).Answer: c≈0.1565.
Problem 2.10
Let f(x)=e3x on [0,3ln4].(a) Check MVT hypotheses.(b) Find c∈(0,3ln4) with f′(c)=3ln4−0f(3ln4)−f(0).Answer: f′(x)=3e3x,f(3ln4)=4,f(0)=1.Average slope =3ln44−1=3ln43=ln49.Solve 3e3c=ln49⇒e3c=ln43⇒c=31ln(ln43).Answer: c=31ln(ln43).
Taylor Series
Problem 3.1
Let f∈Cn+1(I) where I is an open interval containing a and x.Derive the Taylor formula of order n about a with the Lagrange form of the remainder.Outline of proof:1. Let Pn(t)=k=0∑nk!f(k)(a)(t−a)k be the nth Taylor polynomial at a.2. Define M:=(x−a)n+1f(x)−Pn(x)(assuming x=a), and considerΦ(t):=f(t)−Pn(t)−M(t−a)n+1.Then Φ(a)=f(a)−Pn(a)=0 and by the choice of M we have Φ(x)=0.3. Because f∈Cn+1(I) and Pn is a polynomial, Φ∈Cn+1(I).Apply Rolle’s theorem to Φ on [a,x]: there exists c1∈(a,x) with Φ′(c1)=0.Repeat: since Φ′(a)=Φ′(c1)=0, apply Rolle to Φ′ to find c2∈(a,c1) with Φ′′(c2)=0,and continue this process n+1 times.After n+1 applications we obtain a point ξ∈(a,x) (or (x,a) if x<a) such that Φ(n+1)(ξ)=0.4. Compute Φ(n+1)(t)=f(n+1)(t)−(n+1)!M.Hence 0=Φ(n+1)(ξ)=f(n+1)(ξ)−(n+1)!M⇒M=(n+1)!f(n+1)(ξ).5. Substitute M back into the definition of M or into Φ(x)=0 to obtainf(x)=Pn(x)+(n+1)!f(n+1)(ξ)(x−a)n+1for some ξ between a and x.Answer: f(x)=k=0∑nk!f(k)(a)(x−a)k+(n+1)!f(n+1)(ξ)(x−a)n+1,ξ∈(a,x)(or (x,a)).
Problem 3.2
Let f(x)=ln(1+x).Find the third-order Taylor polynomial about a=0 and use it to approximate f(0.5).Give the Lagrange form of the remainder R3(0.5) and produce an explicit numerical bound.Detailed solution:Derivatives: f′(x)=1+x1,f′′(x)=−(1+x)21,f(3)(x)=(1+x)32,f(4)(x)=−(1+x)46.Taylor polynomial of degree 3 about 0:P3(x)=f(0)+f′(0)x+2!f′′(0)x2+3!f(3)(0)x3=0+1⋅x+2−1x2+62x3=x−2x2+3x3.Evaluate at x=21:P3(21)=21−81+241=2412−3+1=2410=125≈0.4166666.Lagrange remainder: R3(x)=4!f(4)(ξ)x4for some ξ∈(0,x).Thus R3(21)=24f(4)(ξ)(21)4=(1+ξ)4−6⋅241⋅161=−24⋅166⋅(1+ξ)41.Magnitude bound: for ξ∈(0,21) we have (1+ξ)≥1⇒∣f(4)(ξ)∣≤6.Hence ∣R3(21)∣≤246⋅(21)4=246⋅161=3846=641=0.015625.Numerical check: f(0.5)=ln(1.5)≈0.4054651081,P3(0.5)=125≈0.4166666667,∣f(0.5)−P3(0.5)∣≈0.0112015586≤0.015625(within the bound).Answer: P3(x)=x−2x2+3x3,P3(21)=125,∣R3(21)∣≤641.
Problem 3.3
Let f(x)=arctanx.Find the fifth-order Taylor polynomial about a=0 and use it to approximate f(0.5).Give the remainder in alternating-series form and a numerical error bound.Answer: P5(x)=x−3x3+5x5.P5(21)=21−241+1601=16080−10+3=16073=0.45625.(Alternating series remainder) ∣R5(0.5)∣≤7(0.5)7=70.0078125≈0.00111607.Numeric: arctan(0.5)≈0.4636476090,error ≈0.0073976090(within the bound above).
Problem 3.4
Let f(x)=lnx.Find the second-order Taylor polynomial about a=1 and use it to approximate f(1.2).Give the Lagrange remainder and a numerical bound.Answer: P2(x)=(x−1)−21(x−1)2.P2(1.2)=0.2−21(0.2)2=0.2−0.02=0.18.Lagrange remainder: R2(1.2)=3!f(3)(ξ)(0.2)3=62⋅ξ3(0.2)3for some ξ∈(1,1.2).Bound using ξ≥1:∣R2(1.2)∣≤62⋅0.008=31⋅0.008=30.008≈0.0026667.Numeric: ln(1.2)≈0.1823215568,error ≈0.0023215568.
Problem 3.5
Let f(x)=1+x1.Find the third-order Taylor polynomial about a=0 and use it to approximate f(0.3).Give the remainder in Lagrange form and a numerical bound.Answer: P3(x)=1−x+x2−x3.P3(0.3)=1−0.3+0.09−0.027=0.763.Lagrange remainder: R3(0.3)=4!f(4)(ξ)(0.3)4 for some ξ∈(0,0.3),f(4)(t)=(1+t)524(−1)4+1=−(1+t)524.Bound using (1+ξ)≥1:∣R3(0.3)∣≤2424⋅(0.3)4=(0.3)4=0.0081.Numeric: 1.31≈0.7692307692,error ≈0.0062307692≤0.0081.
Miscellaneous
Problem 4.1
Show that x3−3ax+2b=0 has: (a) one real root if a<0,and (b) three real roots if b2<a3.Answer: (a)a<0⇒one real root.(b)b2<a3⇒three real roots (all real).
Problem 4.2
Show that x3−tx+1=0 has: (a) one real root if t<0,and (b) three real roots if t3>427(i.e. t>(27/4)1/3).Answer: (a)t<0⇒one real root.(b)t3>427⇒three real roots.
Intermediate Value Theorem
Problem 5.1
Show that the equation sinx=2x has a solution in (2π,π).Detailed solution:Define F(x)=sinx−2x.The function F is continuous on [2π,π].F(2π)=1−4π>0(since π<4),F(π)=0−2π=−2π<0.Because F(2π)>0 and F(π)<0 and F is continuous, the Intermediate Value Theoremguarantees at least one c∈(2π,π) with F(c)=0, i.e. sinc=2c.Answer: ∃c∈(2π,π)such that sinc=2c.
Problem 5.2
Show that f(x)=x5+x−1 has a unique real root, and locate it in an interval of length 0.1.Answer: f(0)=−1<0,f(1)=1>0⇒ at least one root in (0,1).Since f′(x)=5x4+1>0∀x,the function is strictly increasing, so the root is unique. Numerical sign checks give f(0.75)≈−0.0127<0,f(0.8)≈0.12768>0⇒ unique root in (0.75,0.8).
Problem 5.3
Let F(x)=∫0xet2dt(x∈[0,1]).Show there exists c∈(0,1) with F(c)=21F(1).Answer: F is continuous and strictly increasing on [0,1](F(0)=0<F(1)),so by IVT there is a unique c∈(0,1)such that F(c)=21F(1).
Problem 5.4
Let f:[0,1]→[0,1] be continuous. Show that f has a fixed point x∈[0,1](f(x)=x).Answer (proof outline): Considerg(x)=f(x)−x,continuous on [0,1].Since f(x)∈[0,1],g(0)=f(0)−0≥0,g(1)=f(1)−1≤0.Hence g(0)⋅g(1)≤0, so by IVT there exists c∈[0,1] with g(c)=0,which is f(c)=c(a fixed point).
Problem 5.5
Consider pk(x)=x3−3x+k.Show that pk(x)=0 has three distinct real roots precisely when −2<k<2.Answer: pk′(x)=3x2−3=0⇒x=±1.Compute pk(−1)=−1+3+k=k+2,pk(1)=1−3+k=k−2.For three real roots we need the local maximum pk(−1)>0 and the local minimum pk(1)<0, i.e. k+2>0,k−2<0,which is −2<k<2.(If k≤−2 or k≥2 there are fewer than three real roots.)
