<br />For what integral values of x, 2raised to x x 5 raised to x ends in 5?(C) no value of x A 0(B) 1(D) 2 It degree of both f(x) and (f(x) +g (x) is 18, then degree of g(x) must be<br />(A)> 18 (B) equal to 18C less than equal to (D) none of these if the median of the data X1, X, X3, X, X5, X6, X. xs is 'a', then find the median of the data X3 X4, X.X; Where (X) = X1 <X2<X3 < X4< X6 < x7<X8) A a B a\2 C a\4 (D) can not say
Answers
Answer:
2
Theorem 6.3 : (Mean Value Theorem) Let f be 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).
Proof: Let
g(x) = f(x) −
f(b) − f(a)
b − a
(x − a).
Then g(a) = g(b) = f(a). The result follows by applying Rolle’s Theorem to g. ¤
The mean value theorem is an important result in calculus and has some important applications
relating the behaviour of f and f
0
. For example, if we have a property of f
0 and we want to see
the effect of this property on f, we usually try to apply the mean value theorem. Let us see some
examples.
Example 1 : Let f : [a, b] → R be differentiable. Then f is constant if and only if f
0
(x) = 0 for
every x ∈ [a, b].
Proof : Suppose that f is constant, then from the definition of f
0
(x) it is immediate that f
0
(x) = 0
for every x ∈ [a, b].
To prove the converse, let a < x ≤ b. By the mean value theorem there exists c ∈ (a, x) such
that f(x)−f(a) = f
0
(c)(x−a). Since f
0
(c) = 0, we conclude that f(x) = f(a), that is f is constant.
(If we try to prove the converse directly from the definition of f
0
(x) we will be in trouble.) ¤
Example 2 : Suppose f is continuous on [a, b] and differentiable on (a, b).
(i) If f
0
(x) 6= 0 for all x ∈ (a, b), then f is one-one (i.e, f(x) 6= f(y) whenever x 6= y).
(ii) If f
0
(x) ≥ 0 (resp. f
0
(x) > 0) for all x ∈ (a, b) then f is increasing (resp. strictly increasing)
on [a, b]. (We have a similar result for decreasing functions.)
Proof : Apply the mean value theorem as we did in the previous example. (Note that f can be
one-one but f
0
can be 0 at some point, for example take f(x) = x
3 and x = 0.)
Problem 3 : Use the mean value theorem to prove that | sinx−siny | ≤ | x−y | for all x, y ∈ R.
Solution : Let x, y ∈ R. By the mean value theorem sinx−siny = cosc (x−y) for some c between
x and y. Hence | sinx − siny | ≤ | x − y |.
Problem 4 : Let f be twice differentiable on [0, 2]. Show that if f(0) = 0, f(1) = 2 and f(2) = 4,
then there is x0 ∈ (0, 2) such that f
00(x0) = 0.
Solution : By the mean value theorem there exist x1 ∈ (0, 1) and x2 ∈ (1, 2) such that
f
0
(x1) = f(1) − f(0) = 2 and f
0
(x2) = f(2) − f(1) = 2.
Apply Rolle’s theorem to f
0 on [x1, x2].
Problem 5 : Let a > 0 and f : [−a, a] → R be continuous. Suppose f
0
(x) exists and f
0
(x) ≤ 1 for
all x ∈ (−a, a). If f(a) = a and f(−a) = −a, then show that f(x) = x for every x ∈ (−a, a).
Solution : Let g(x) = f(x)−x on [−a, a]. Note that g
0
(x) ≤ 0 on (−a, a). Therefore, g is decreasing.
Since g(a) = g(−a) = 0, we have g = 0.
This problem can also be solved by applying the MVT for g on [−a, x] and [x, a].