Unit III Practice Questions
Calculus for Computer Science
I B.Tech I Semester
1. Rolles Theorem
Statement: If a function f (x) is continuous on [a, b], differentiable on (a, b) and f (a) =
f (b), then there exists at least one c ∈ (a, b) such that
f 0 (c) = 0.
1 1
(i) f (x) = x + on ,2
x 2
1
6 0. Since 0 ∈
Step 1: The function is continuous and differentiable for x = / , 2 , the
2
conditions are satisfied.
Step 2:
1 1 5 1 5
f = +2= , f (2) = 2 + =
2 2 2 2 2
Step 3:
1
f 0 (x) = 1 −
x2
Step 4:
1
1− 2
= 0 ⇒ c2 = 1 ⇒ c = 1
c
c=1
x3
(ii)f (x) = − 3x on [−3, 3]
3
Verify Rolles Theorem for the function
x3
f (x) = − 3x on [−3, 3].
3
Step 1: Continuity
The given function
x3
f (x) = − 3x
3
is a polynomial. Hence, it is continuous on the closed interval [−3, 3].
1
,Step 2: Differentiability
Since f (x) is a polynomial, it is differentiable on the open interval (−3, 3).
Step 3: Verify the end point values
(−3)3 −27
f (−3) = − 3(−3) = + 9 = −9 + 9 = 0,
3 3
(3)3 27
f (3) = − 3(3) = − 9 = 9 − 9 = 0.
3 3
Thus,
f (−3) = f (3).
Step 4: Find the derivative
f 0 (x) = x2 − 3
Step 5: Solve f 0 (c) = 0
√
x2 − 3 = 0 ⇒ x2 = 3 ⇒ x = ± 3
Step 6: Verification
Since √ √
− 3, 3 ∈ (−3, 3),
all the hypotheses of Rolles Theorem are satisfied.
√
Rolles Theorem is verified for c = ± 3.
2. Lagranges Mean Value Theorem
Statement: If f (x) is continuous on [a, b] and differentiable on (a, b), then there exists
c ∈ (a, b) such that
f (b) − f (a)
f 0 (c) = .
b−a
(i) f (x) = x3 − 2x on [−1, 2]
f (2) = 4, f (−1) = 1
f (2) − f (−1)
=1
2 − (−1)
f 0 (x) = 3x2 − 2
3c2 − 2 = 1 ⇒ c2 = 1
c = ±1
2
, (ii) f (x) = ln(x − 1) on [2, 4]
ln 3 − ln 1 ln 3
=
2 2
1 1 ln 3
f 0 (x) = ⇒ =
x c 2
2
c=
ln 3
1
(iii) f (x) = on [a, b]
x
1
f 0 (x) = −
x2
1 1
− a1 √
b
− 2
= ⇒ c = ab
c b−a
(iv) f (x) = x2 on [a, b]
b 2 − a2
f 0 (x) = 2x ⇒ 2c = =a+b
b−a
a+b
c=
2
(v) Piecewise function
(
x3 , −2 ≤ x ≤ 0,
g(x) =
x2 , 0 < x ≤ 2.
The interval is [−2, 2].
Apply Lagranges Mean Value Theorem
g(2) − g(−2) 4 − (−8) 12
= = = 3.
2 − (−2) 4 4
So we must find c such that
g 0 (c) = 3.
Case 1: −2 < c < 0
3c2 = 3 ⇒ c2 = 1 ⇒ c = −1.
3
Calculus for Computer Science
I B.Tech I Semester
1. Rolles Theorem
Statement: If a function f (x) is continuous on [a, b], differentiable on (a, b) and f (a) =
f (b), then there exists at least one c ∈ (a, b) such that
f 0 (c) = 0.
1 1
(i) f (x) = x + on ,2
x 2
1
6 0. Since 0 ∈
Step 1: The function is continuous and differentiable for x = / , 2 , the
2
conditions are satisfied.
Step 2:
1 1 5 1 5
f = +2= , f (2) = 2 + =
2 2 2 2 2
Step 3:
1
f 0 (x) = 1 −
x2
Step 4:
1
1− 2
= 0 ⇒ c2 = 1 ⇒ c = 1
c
c=1
x3
(ii)f (x) = − 3x on [−3, 3]
3
Verify Rolles Theorem for the function
x3
f (x) = − 3x on [−3, 3].
3
Step 1: Continuity
The given function
x3
f (x) = − 3x
3
is a polynomial. Hence, it is continuous on the closed interval [−3, 3].
1
,Step 2: Differentiability
Since f (x) is a polynomial, it is differentiable on the open interval (−3, 3).
Step 3: Verify the end point values
(−3)3 −27
f (−3) = − 3(−3) = + 9 = −9 + 9 = 0,
3 3
(3)3 27
f (3) = − 3(3) = − 9 = 9 − 9 = 0.
3 3
Thus,
f (−3) = f (3).
Step 4: Find the derivative
f 0 (x) = x2 − 3
Step 5: Solve f 0 (c) = 0
√
x2 − 3 = 0 ⇒ x2 = 3 ⇒ x = ± 3
Step 6: Verification
Since √ √
− 3, 3 ∈ (−3, 3),
all the hypotheses of Rolles Theorem are satisfied.
√
Rolles Theorem is verified for c = ± 3.
2. Lagranges Mean Value Theorem
Statement: If f (x) is continuous on [a, b] and differentiable on (a, b), then there exists
c ∈ (a, b) such that
f (b) − f (a)
f 0 (c) = .
b−a
(i) f (x) = x3 − 2x on [−1, 2]
f (2) = 4, f (−1) = 1
f (2) − f (−1)
=1
2 − (−1)
f 0 (x) = 3x2 − 2
3c2 − 2 = 1 ⇒ c2 = 1
c = ±1
2
, (ii) f (x) = ln(x − 1) on [2, 4]
ln 3 − ln 1 ln 3
=
2 2
1 1 ln 3
f 0 (x) = ⇒ =
x c 2
2
c=
ln 3
1
(iii) f (x) = on [a, b]
x
1
f 0 (x) = −
x2
1 1
− a1 √
b
− 2
= ⇒ c = ab
c b−a
(iv) f (x) = x2 on [a, b]
b 2 − a2
f 0 (x) = 2x ⇒ 2c = =a+b
b−a
a+b
c=
2
(v) Piecewise function
(
x3 , −2 ≤ x ≤ 0,
g(x) =
x2 , 0 < x ≤ 2.
The interval is [−2, 2].
Apply Lagranges Mean Value Theorem
g(2) − g(−2) 4 − (−8) 12
= = = 3.
2 − (−2) 4 4
So we must find c such that
g 0 (c) = 3.
Case 1: −2 < c < 0
3c2 = 3 ⇒ c2 = 1 ⇒ c = −1.
3
