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f ( b ) f ( a ) b a 0 .

Since f is a differentiable function, by the Mean Value Theorem, there exists c ( a , b ) such that

f ( c ) = f ( b ) f ( a ) b a .

Therefore, there exists c I such that f ( c ) 0 , which contradicts the assumption that f ( x ) = 0 for all x I .

From [link] , it follows that if two functions have the same derivative, they differ by, at most, a constant.

Corollary 2: constant difference theorem

If f and g are differentiable over an interval I and f ( x ) = g ( x ) for all x I , then f ( x ) = g ( x ) + C for some constant C .

Proof

Let h ( x ) = f ( x ) g ( x ) . Then, h ( x ) = f ( x ) g ( x ) = 0 for all x I . By Corollary 1, there is a constant C such that h ( x ) = C for all x I . Therefore, f ( x ) = g ( x ) + C for all x I .

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function f is increasing over I if f ( x 1 ) < f ( x 2 ) whenever x 1 < x 2 , whereas f is decreasing over I if f ( x ) 1 > f ( x 2 ) whenever x 1 < x 2 . Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing ( [link] ). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.

This fact is important because it means that for a given function f , if there exists a function F such that F ( x ) = f ( x ) ; then, the only other functions that have a derivative equal to f are F ( x ) + C for some constant C . We discuss this result in more detail later in the chapter.

A vaguely sinusoidal function f(x) is graphed. It increases from somewhere in the second quadrant to (a, f(a)). In this section it is noted that f’ > 0. Then in decreases from (a, f(a)) to (b, f(b)). In this section it is noted that f’ < 0. Finally, it increases to the right of (b, f(b)) and it is noted in this section that f’ > 0.
If a function has a positive derivative over some interval I , then the function increases over that interval I ; if the derivative is negative over some interval I , then the function decreases over that interval I .

Corollary 3: increasing and decreasing functions

Let f be continuous over the closed interval [ a , b ] and differentiable over the open interval ( a , b ) .

  1. If f ( x ) > 0 for all x ( a , b ) , then f is an increasing function over [ a , b ] .
  2. If f ( x ) < 0 for all x ( a , b ) , then f is a decreasing function over [ a , b ] .

Proof

We will prove i.; the proof of ii. is similar. Suppose f is not an increasing function on I . Then there exist a and b in I such that a < b , but f ( a ) f ( b ) . Since f is a differentiable function over I , by the Mean Value Theorem there exists c ( a , b ) such that

f ( c ) = f ( b ) f ( a ) b a .

Since f ( a ) f ( b ) , we know that f ( b ) f ( a ) 0 . Also, a < b tells us that b a > 0 . We conclude that

f ( c ) = f ( b ) f ( a ) b a 0 .

However, f ( x ) > 0 for all x I . This is a contradiction, and therefore f must be an increasing function over I .

Key concepts

  • If f is continuous over [ a , b ] and differentiable over ( a , b ) and f ( a ) = 0 = f ( b ) , then there exists a point c ( a , b ) such that f ( c ) = 0 . This is Rolle’s theorem.
  • If f is continuous over [ a , b ] and differentiable over ( a , b ) , then there exists a point c ( a , b ) such that
    f ( c ) = f ( b ) f ( a ) b a .

    This is the Mean Value Theorem.
  • If f ( x ) = 0 over an interval I , then f is constant over I .
  • If two differentiable functions f and g satisfy f ( x ) = g ( x ) over I , then f ( x ) = g ( x ) + C for some constant C .
  • If f ( x ) > 0 over an interval I , then f is increasing over I . If f ( x ) < 0 over I , then f is decreasing over I .

Questions & Answers

How to do basic integrals
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ram Reply
find the integral of tan tanxdx
Lateef Reply
-ln|cosx| + C
Jug
discuss continuity of x-[x] at [ _1 1]
Atshdr Reply
Given that u = tan–¹(y/x), show that d²u/dx² + d²u/dy²=0
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find the limiting value of 5n-3÷2n-7
Joy Reply
Use the first principal to solve the following questions 5x-1
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175000/9*100-100+164294/9*100-100*4
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mode of (x+4) is equal to 10..graph it how?
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66
ram
6
ram
6
Cajab
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nelson
integrals of 1/6-6x-5x²
Namwandi Reply
derivative of (-x^3+1)%x^2
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(-x^5+x^2)/100
Sarada
(-5x^4+2x)/100
Sarada
oh sorry it's (-x^3+1)÷x^2
Misha
-5x^4+2x
Sarada
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Sarada
find the derivative of the following y=4^e5x y=Cos^2 y=x^inx , x>0 y= 1+x^2/1-x^2 y=Sin ^2 3x + Cos^2 3x please guys I need answer and solutions
Ga Reply
differentiate y=(3x-2)^2(2x^2+5) and simplify the result
Ga
72x³-72x²+106x-60
okhiria
y= (2x^2+5)(3x+9)^2
lemmor
solve for dy/dx of y= 8x^3+5x^2-x+5
Ga Reply
192x^2+50x-1
Daniel
are you sure? my answer is 24x^2+10x-1 but I'm not sure about my answer .. what do you think?
Ga
24x²+10x-1
Eyad
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y= (2x^2+5)(3x+9)^2
lemmor
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Fernando
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Jug
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yhin
of coursr
okhiria
but i think, it's more complicated than calculus 1
Jug
Hello can someone help me with calculus one...
Jainaba
find the derivative of y= (2x+3)raise to 2 sorry I didn't know how to put the raise correctly
Ga Reply
8x+12
Dhruv
8x+3
okhiria
d the derivative of y= e raised to power x
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rates of change and tangents to curves
Kyaw Reply
how can find differential Calculus
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Practice Key Terms 2

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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