Expertise of University have researched and collected Study Material Mean Value Theorems study material in pdf format. This pdf formatted University Study Material Mean Value Theorems covered the subject concepts in a comprehend manner for easy reference to their students.
University
UniversityCourse
Study Materialwww.akhiai. www.akhiai. MEAN VALUE THEOREMS Rolle'sTheorem : If a function f : [a, b] →R is such that i) It is continuous on [a, b]
ii) It is derivable on (a, b) and iii) f(a) = f(b) then there exists at least one (,)ca b∈ such that ()0f c′=.
Lagrange's mean -value theorem or first mean - value theorem :
If a function f : [a, b] →R is such that i) It is continuous on [a, b].
ii)It is derivable on (a, b) then there exists at least one (,)ca b∈ such that( ) ( )()f b f afcb a−′=− VSAQ’S 1.Verify Rolle’s theorem for the function x2 – 1 on [–1, 1].
Sol. Let f(x) = x2 – 1 f is continuous on [–1, 1] and f is differentiable on (–1, 1) since f(–1) = f(1) = 0 and ∴ By Rolle’s theoremc(1,1)∃ ∈ − Such that f′(c) = 0
f′(x) = 2x = 0 ∴ = f′(c) = 0 2c = 0 ⇒ c = 0 The point c = 0 ∈ (–1, 1) Then Rolle’s theorem is verified.
2. sin x – sin 2x on [0, π].
Sol. Let f(x) = sin x – sin 2x f is continuous on [0, π] f is differentiable on (0, π) since f(0) = f(π) = 0 and ∴ By Rolle’s theoremc(0,)∃ ∈ π www.akhiai.
www.akhiai. Such that f′(c) = 0 f′(x) = cos x – 2 cos 2x f′(c) = 0 ⇒cos c – 2 cos 2x = 0 ⇒cos c – 2(2cos2 c – 1) = 0 cos c – 4 cos2 c + 2 = 0
4cos2 c – cos c – 2 = 0 11 1 32 1 33cosc881 33c cos8−± + ±= = ±∴= 3. log(x2 + 2) – log 3 on [–1, 1].
Sol. Let f(x) = log (x2 + 2) – log 3 f is continuous on [–1, 1] and f is differentiable on (–1, 1) since f(–1) = f(1) = 0 ∴ By Rolle’s theoremc(1,1)∃ ∈ − Such that f′(c) = 0 f′(x) = 21x2+(2x)
f′(c) = 22cc2+ = 0 2c = 0 ⇒ c = 0 The point c = 0 ∈ (–1, 1)4. It is given that Rolle’s theorem holds for the function f(x) = x3 + bx2 + ax on [1, 3] with1c 2t3= + . Find the values of a and b.
Sol. Given f(x) = x3 + bx2 + ax www.akhiai. www.akhiai. 222
222f (x) 3x 2bx af'(x)0 3c 2bc a 02b 4b 12ac6b b 3ac3
1 b b 3a23 3b b 3a 12 and3 3 3′= + +∴=⇔+ + =−±−⇔=−±−=
−±−+ =− −= = 2b 6 and b 3a 336 3 3a 33 3a a 11⇔=−=⇒−=⇒=⇒= Hence a = 11, b = –6.
5. Find a point on the graph of the curve y = x3, where the tangent is parallel to the chord joining (1, 1) and (3, 27).
Sol. Given Points (1, 1) and (3, 27).
Slope of chord = 27 1133 1−=−Given y = x3 222 2dy3xdx
Slope 3x1313 3x x313 39x33=⇒== ⇒=⇒= =
3339 39 39 13 39y x327 9 = = = = ∴ The point on the curve is 39 13 39,3 9 6. Find ‘c’, so that f(b)f(a)f (c)ba−′=− in the following case. f(x) = x2 – 3x – 1, a = –11/7,
b = 13/7.
Sol. ( )13 169 3(13)fb f 17 49 7 = =− − www.akhiai. www.akhiai. 169 273 49 15349 49− − −= 11 121 3(11)f(a)f 17 49 7121 231 49 303
49 49f (x) 2x 3f (c) 2c 3− − = =− − +−= =′=−
′=− Given f(b)f(a)f (c)ba−′=− 153 303 45649 49 492c 313 11 2477 7456 7 192c 349 24 719 2 12c 3 c77 7− −−
−= =+− −−= × =−= + =⇒=7. Verify the Rolle’s theorem for the function (x2 – 1)(x – 2) on [–1, 2]. Find the point in
the interval where the derivate vanishes.
Sol. Let f(x) = (x2 – 1)(x – 2) = x3 – 2x2 – x + 2 f is continuous on [–1, 2] since f(–1) = f(2) = 0 and f is differentiable on (–1, 2) ∴ By Rolle’s theoremc(1,2)∃ ∈ −
Let f′(c) = 0 f′(x) = 3x2 – 4x – 1 3c2 – 4c – 1 = 0 416 12 4 28c6 6± + ±= = 27c3±⇒=
8. Verify the conditions of the Lagrange’s mean value theorem for the following function. In each case find a point ‘c’ in the interval as stated by the theorem.
sin x – sin 2x on [0, π].
Sol. Let f(x) = sin x – sin 2x www.akhiai. www.akhiai. f is continuous on [0, π] and f is differentiable on (0,π) Given f(x) = sinx – sin 2x
f′(x) = cos x – 2 cos 2x By Lagrange’s mean value than c(0,)∃ ∈ π such there 2 22
1f ( ) f (0)f (c) cos c 2cos 2c 00cos c 2(2cos c 1) 0 cos c 4cos c 2 01 1 32 1 334cos c cos c 2 0 cos c881 33c cos8−π −′=⇒−=π−⇒− −=⇒−+ =
± + ±⇒− −=⇒= = ±⇒=