Analysis of Functions IIncrease, Decrease, and Concavity

increasing

increasing

decreasing constant

Definition. Let f(x) be defined on an interval, and let x1 and x2denote points in that interval.(a) f is increasing on the interval if f(x1) < f(x2) whenever x1< x2.(b) f is decreasing on the interval if f(x1) > f(x2) whenever x1< x2.(c) f is constant on the interval if f(x1) = f(x2) for all x1and x2.

Here are some other examples.

Increasing functions Decreasing functions

Notice that the shape of the functions can vary.

Theorem. Let f be a function that is continuous on a closed interval [a, b], and differentiable on the open interval (a, b).(a) If f ′(x) > 0 for every value of x in (a, b), then f is increasing in [a, b].(b) If f ′(x) < 0 for every value of x in (a, b), then f is decreasing in [a, b].(c) If f ′(x) = 0 for every value of x in (a, b), then f is constant on [a, b].

Positive derivative = increasing

Negative derivative = decreasing

Zero derivative = constant

Example. Let

Find the intervals on which f is increasing and the intervals on which f is decreasing.

2( ) 2 1f x x x= − +

Solution. ( ) 2 2 2( 1)f x x x′ = − = −

It is clear that this derivative is positive when x > 1 and negative when x < 1. Thus the function is decreasing in the interval (−∞, 1) and is increasing in the interval (1, ∞).

The graph confirms this analysis.

Example. Let

Find the intervals on which f is increasing and the intervals on which f is decreasing.

4 2( ) 8 16f x x x= − +

Solution. 3 3( ) 4 16 4( 4 )2 4 ( 4) 4 ( 2)( 2).

f x x x x xx x x x x

′ = − = −= − = − +

If x < −2, all three factors are negative, so the product is negative. If −2< x <0, then x + 2 is positive, while x and x −2 are negative. Thus the product is positive. If 0 < x < 2, then x −2 is negative, while x and x + 2 are positive, so again the product is negative.If x > 2, then all factors are positive, and so the product is positive

This is the graph of the function, which confirms the analysis.

Concavity

Concavity refers to the shape of a curve, rather than its direction. A curve is concave up or down, if it will “hold water, or “spill water”.

Concave up Concave Downincreasing – decreasing increasing – decreasing

Definition. If f is differentiable on an open interval I, then f is said to be concave up on I if f′ is increasing on I, and f is said to be concave down on I if f′ is decreasing on I.

We know that a function is increasing or decreasing according to whether its derivative is positive or negative. This result can be applied to the function f′ if that function has a derivative; in other words if f has a second derivative on I. This leads to the following useful theorem.

Theorem. Let f be twice differentiable on an open interval I. (a) If f′′ (x) > 0 on I, then f is concave up on I.(b) If f′′ (x) < 0 on I, then f is concave down on I.

+++

−−−

Concave up – holds waterSecond derivative > 0

Concave down – spills waterSecond derivative < 0

+

++ −

−

−

Example. Find the open intervals on which the function

is concave up, and the open intervals on which it is concave down.

3 2( ) 2 1f x x x x= + − −

Solution. The derivative of f is Thus the second derivative is

23 2 2.x x+ −6 2 2(3 1).x x+ = +

This means that f is concave down if 12(3 1) 0 or .3

x x+ < <−

Similarly, f is concave up if 1 .3

x>−

3 2( ) 2 1f x x x x= + − −

Concave Down in this half

Concave Up in this half

3 2( ) 2 1f x x x x= + − −

Definition. If f is continuous on an open interval containing the point x0, and if f changes the direction of its concavity at that point, then we say that f has an inflection point at x0, and we call the point (x0, f(x0)) on the graph of f an inflection point of f.

Inflection point

3 2( ) 2 1f x x x x= + − −

Example. Find the inflection points of and , and confirm the results by looking at the graphs of these functions.

xxe cos( )x

Solution. ( )2 2 .x x x x x xxe e xe e xe e x′′ ′ = + = + = +

xxe

cos( ) ( sin( )) cos( ).x x x′′ ′= − =−

Thus the cosine is concave up where it is negative and concave down where it is positive. Thus the inflection points come where cos (x) is 0.

Since the exponential is never negative or 0, we see that is concave down if x < −2 and concave up if x > −2. Thus −2 is an inflection point.

The graphs of these two functions confirm these results.

Example. Use the graph of the function y = f(x), shown below,

to determine the signs of at the points A, B, and C.2

and 2

dy d ydx dx

A

B

C

Solution. At A, 2

>0 and 02

dy d ydx dx

<

2At , <0 and 0

2dy d yBdx dx

<2

At , >0 and 02

dy d yCdx dx

>

( )f x

Example. Use the graph of the derivative f′ (x), shown below,

to determine the signs of at the points A, B, and C.2

and 2

dy d ydx dx

A

B

C

Solution. At A, 2

>0 and 02

dy d ydx dx

>

2At , >0 and 0

2dy d yBdx dx

<2

At , <0 and 02

dy d yCdx dx

>

( )f x′

Example. Use the graph of y = f(x), shown below to identify all intervals where the function is increasing, decreasing, concave up, concave down. Find all values of x at which f has an inflection point.

1 23 4 5

6 7

Solution. The function f is increasing on the closed intervals [1, 2], and [4, 7]. It is decreasing on the closed interval [2, 4].

The function f is concave up on the open intervals (3, 5) and (6, 7). It is concave down on the open intervals (1, 3) and (5, 6).

1 23 4 5

6 7

Example. For the function find all intervals on which f is increasing, all intervals on which f is decreasing, all open intervals on which f is concave up, all intervals on which f is concave down, and all points of inflection of f.

3( ) 5 12f x x x= + −

Thus the function is increasing in the interval (−2 , 2), and decreasing in intervals (−∞ , −2) and (2, ∞).

It is concave up in the interval (−∞ , 0) and concave down in the interval (0, −∞ ).

Solution. 2 2( ) 12 3 3(4 )f x x x′ = − = − and ( ) 6 .f x x′′ =−From this it is clear that f′ (x) > 0 if and f′ (x) < 0 2 4x <

We also have f′′ (x) > 0 if x < 0 and f′′ (x) < 0 if x > 0. 2if 4.x >

Here is the graph of the function.

Example. For the function find all intervals on which f is increasing, all intervals on which f is decreasing, all open intervals on which f is concave up, all intervals on which f is concave down, and all points of inflection of f.

4 3( ) 3 4 2f x x x= − +

Solution.

Thus the function is increasing in the intervals (1, ∞) and the function is decreasing in interval (−∞ , 1) .

It is concave up in the intervals (−∞ , 0) and (2/3, ∞), and it is concave down in the interval (0, 2/3 ).

and

From this it is clear that f′ (x) > 0 if x > 1

if 1.x<

and f′ (x) < 0 We also have f′′ (x) > 0 if x < 0 and if x > 2/3, while

2( ) 36 24 12 (3 2).f x x x x x′′ = − = −3 2 2( ) 12 12 12 ( 1)f x x x x x′ = − = −

f′′(x) < 0 if 0 < x < 2/3.

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At x equal 0, 2/3, and 1, either the direction or the curvature changes. We plot these points on the graph, and divide the planeinto regions.

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f ′ (x) < 0f ′′ (x) > 0 f ′ (x) < 0

f ′′ (x) < 0f ′ (x) < 0f ′′ (x) > 0

f ′ (x) > 0f ′′ (x) > 0

Recall the characterization of the pieces of a smooth curve

Concave up Concave Downincreasing – decreasing increasing – decreasing

1. f ′ > 0, f ′′ > 0 2. f ′ < 0, f ′′ > 03. f ′ > 0, f ′′ < 0 4. f ′ < 0, f ′′ < 0

1 2 3 4

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f ′ (x) < 0f ′′ (x) < 0

f ′ (x) < 0f ′′ (x) > 0

f ′ (x) > 0f ′′ (x) > 0

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f ′ (x) < 0f ′′ (x) > 0

f ′ (x) > 0f ′′ (x) > 0

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f ′ (x) > 0f ′′ (x) > 0

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Here is the actual graph.