5.If f is the derivative of some function on [a,b],then there exists a number c€(a,b) such that Integral(fdx)=
1) f(c)
2) f(c).(b-a)
3) f(c).(b+a)
4) (b+a-c)f(c)
Answers
Answer:
Step-by-step explanation:
The reader must be familiar with the classical maxima and minima problems from calculus. For
example, the graph of a differentiable function has a horizontal tangent at a maximum or minimum
point. This is not quite accurate as we will see.
Definition : Let f : I → R, I an interval. A point x0 ∈ I is a local maximum of f if there is
a δ > 0 such that f(x) ≤ f(x0) whenever x ∈ I ∩ (x0 − δ, x0 + δ). Similarly, we can define local
minimum.
Theorem 6.1 : Suppose f : [a, b] → R and suppose f has either a local maximum or a local
minimum at x0 ∈ (a, b). If f is differentiable at x0 then f
0
(x0) = 0.
Proof: Suppose f has a local maximum at x0 ∈ (a, b). For small (enough) h, f(x0 + h) ≤ f(x0).
If h > 0 then
f(x0 + h) − f(x0)
h
≤ 0.
Similarly, if h < 0, then
f(x0 + h) − f(x0)
h
≥ 0.
By elementary properties of the limit, it follows that f
0
(x0) = 0. ¤
We remark that the previous theorem is not valid if x0 is a or b. For example, if we consider the
function f : [0, 1] → R such that f(x) = x, then f has maximum at 1 but f
0
(x) = 1 for all x ∈ [0, 1].
The following theorem is known as Rolle’s theorem which is an application of the previous
theorem.
Theorem 6.2 : Let f be continuous on [a, b], a < b, and differentiable on (a, b). Suppose f(a) =
f(b). Then there exists c such that c ∈ (a, b) and f
0
(c) = 0.
Proof: If f is constant on [a, b] then f
0
(c) = 0 for all c ∈ [a, b]. Suppose there exists x ∈ (a, b) such
that f(x) > f(a). (A similar argument can be given if f(x) < f(a)). Then there exists c ∈ (a, b)
such that f(c) is a maximum. Hence by the previous theorem, we have f
0
(c) = 0. ¤
Problem 1 : Show that the equation x
13 + 7x
3 − 5 = 0 has exactly one (real) root.
Solution : Let f(x) = x
13 + 7x
3 − 5. Then f(0) < 0 and f(1) > 0. By the IVP there is at least one
positive root of f(x) = 0. If there are two distinct positive roots, then by Rolle’s theorem there is
some x0 > 0 such that f
0
(x0