   Chapter 4.3, Problem 69E

Chapter
Section
Textbook Problem

# Justify (3) for the case h < 0 .

To determine

To justify:

The equation   f(u)gx+h-gxhf(v), for (h<0)

Explanation

1) Concept:

The Fundamental Theorem of Calculus (Part 1):

If f is continuous on [a, b], then the function g is defined by

gx= axftdt    axb

is continuous on [a, b] and differentiable on (a, b) and g'x=fx.

2) Formula:

i) Property of definite integral:

If  mfxM for  axb, then,

m(b-a)abfx dxM(b-a) ii) gx+h-g(x)h=1hxx+hftdt

3) Calculations:

Suppose   h<0. Since f is continuous on x+h, x, by the Extreme Value Theorem there are numbers u and v in x+h, x, such that fu=m  and fv=M which are the minimum and maximum values of f  on x+h, x

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