# Theorems, Derivatives, and Antiderivatives

- StudyBlue
- Canada
- University of Manitoba
- Mathematics
- Mathematics Math 1500
- Robert Borgersen
- Theorems, Derivatives, and Antiderivatives

**Created:**2017-01-08

**Last Modified:**2017-01-09

According to the definition of a decreasing function, we have to show that f(x1)<f(x2)

Because we are given that f '(x) <0, we know that f is differentiable on the interval and continuous on [x1,x2], so by the MVT, there must be a point c, x1<c<x2

f '(c)= f(x2)-f(x1) / x2 - x1

Since f '(c) < 0, the fraction on the right is

negative

Since x1>x2, x2-x1> or = 0, and so f(x2)-f(x1) <0. Thus f(x1)>f(x2), and so the function is decreasing

^{x}

^{x}

^{n}

^{n+1}/ n+1

_{a}x

^{x}) derivative

^{2}x

^{2}x

- f is continuous on [a,b]

-f is differentiable on (a,b)

There is a number c in (a,b) such that f'(c)= f(b)-f(a)/b-a

_{1}and x

_{2}be any two numbers in the interval x

_{1}<x

_{2}. According to definition we must show that f(x

_{2})>f(x

_{1}).

Because we are given that f '(x)>0, we know that f is differentiable on the interval [x

_{1},x

_{2}] and so by the MVT there is a number c, x

_{1}<c<x

_{2}, such that

f '(c) = f(x

_{2})-f(x

_{1}) / x

_{2}-x

_{1}

Since f '(c) >0, the function on the right is positive. Since x

_{2}>x

_{1}, x

_{2}-x

_{1}> or = 0, and so f(x

_{2})>f(x

_{1}), and so the function is increasing.

Let x

_{1}and x

_{2}be any two numbers in (a,b) with x

_{1}< x

_{2}.

Since f is differentiable on (a,b), it must be differentiable on (x

_{1},x

_{2}) and continuous on [x

_{1},x

_{2}].

By applying the MVT to f on [x

_{1},x

_{2}], we get a number c such that x

_{1}<c<x

_{2}and

f'(c)= f(x

_{2})-f(x

_{1}) / x

_{2}-x

_{1}

But we assumed that f'(x)=0 for all x in (a,b) therefore, f(x

_{2})-f(x

_{1})=0 and so f(x

_{2})=f(x

_{1}).

Therefore for any two numbers x

_{1}and x

_{2}in (a,b), f(x

_{1})=f(x

_{2}) and the function is constant on (a,b)

Because we are given that f '(x)< 0, we know that f is differentiable on the interval and specifically on [x1,x2] and so by the MVT there is a point c, x1<c<x2

Such that f '(x)= f(x2)-f(x1) / x2 - x1

Since f'(c)<0, the fraction on the right is negative. Since x2>x1, x2-x1> or = 0, and so f(x2)-f(x1)<0. Thus f(x1)>f(x2), and so

the function is decreasing.

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