#### To determine

**To justify:**

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

#### Answer

The equation f(u)≤gx+h-gxh≤f(v) is justified, 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 a≤x≤b

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

**2) Formula:**

i) Property of definite integral:

If m≤fx≤M for a≤x≤b, then,

m(b-a)≤∫abfx dx≤M(b-a) ii) gx+h-g(x)h=1h∫xx+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.

Then by using formula 1),

mx-x+h ≤ ∫x+hxftdt ≤M(x-x+h)

After simplifying above equation,

m-h ≤ ∫x+hxftdt ≤ M-h

Since fu=m and fv=M, and using property of integration above equation becomes,

fu-h≤ -∫xx+hftdt ≤fv-h

Dividing by -h to above inequality, it becomes,

fu≤1h∫xx+hftdt ≤fv

Now, use formula 2) to replace the middle part of the above inequality

fu≤ gx+h-gxh ≤fv

Hence the equation is justified.

**Conclusion:**

The equation fu≤ gx+h-gxh ≤fv is justified for h<0.