How to prove if a function is increasing, using derivatives?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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iff is not misspelling. In math "iff" means , "if and only if".

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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The function is increasing on an interval iff f'(x)> 0

The function decreasing on an interval iff f'(x)<0

For example:

f(x)= 2x^3-3x^2-12x+1

f'(x)=6x^2-6x-12 = 6(x-2)(x+1)

now f'(x)>0, if x<-1 .... then the function is increasing on (-if,-1)

f'(x)<0 , when -1<x<2... then f is decreasing on (-1,2)

f'(x)>0, when c>2 ... then f is increasing on (2,inf)

 

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First, we'll choose 2 numbers: x1<x2.

Based on the rule of an increasing function, we'll have to prove that f(x1)<f(x2), also.

Supposing that is given the fact that f'(x)>0 and from here, we'll conclude that f is differentiable on the interval [x1, x2].

Now, we'll apply the Mean Value Theorem, which states that:

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

From enunciation, we have that f'(c)>0 and, because we've choosen that x2>x1, we'll havex2-x1>0.

So, the product f'(c)(x2-x1)>0.

But the product f'(c)(x2-x1) = f(x2)-f(x1)

From here, we conclude that f(x2)-f(x1)>0

That means that f(x2)>f(x1), which means that f(x) is an increasing function.

Wiggin42's profile pic

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

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Take the derivative of a function and set it equal to zero. For all the points of x where the derivative equals zero; you have a critical point. Chart this value on a number line. 

So lets say you critical points are  -1 and 5

<---- -1 -------- 5 ------->  

Figure out what the slope is doing during the intervals between the critical points by plugging in a test value into the derivative equation. Wherever the derivative is positive in these intervals, the function is increasing for those intervals. 

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neela | High School Teacher | (Level 3) Valedictorian

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A function f(x) is said to be incresing , if for any b-a> 0, f(b)-f(a) > 0. Or

[f(b) -f(a)]/(b-a) > 0. Or

[f(x+h)-f(x)]/{(x+h)-x)} > 0 for any h>0. Taking limits as h --> 0, we get:

  f'(x) > 0. So   at any point x= c, if f'(c) > 0, then f(x) is said to be increasing at the point x= c as x increases.

Similarly, at  x= c, if f' (c) < 0, then f(x) is said to be decresing at x= c, as x increase.

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neela | High School Teacher | (Level 3) Valedictorian

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A function is said to be increasing if for any (b-a) implies f(b) -f(a)  > 0. Or

[f(x+h) - f(x)}/[(x+h)-x] > 0, for any h.

Or

Taking limits as h--> 0,

f'(x) > 0 is the condtion that at the point x,  f(x) increases with x.

Similarly a function f(x) decreses  at at the point x , f(x) decreases as x increases.

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