Hands first, words secondDrag the two points x1 and x2 along the graph. Inside the shaded interval, is the pair always “falling”, however you place them? Widen the interval and find a “rising” pair.
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Try this
Put the two points anywhere in [−1,1]. What sign does the difference quotient x2−x1f(x2)−f(x1) always have?
Drag both points to the right of x=1. Does “This pair says” change?
Look at the factor x12+x1x2+x22−3 on the right. With both points in [−1,1], can it ever become positive?
Switch to x1 and take x1=−1, x2=1. What does this pair say?
Let the domain of f(x) contain an interval I. If for any two points x1<x2 of I we have f(x1)<f(x2), then f(x) is increasing on I; if we always have f(x1)>f(x2), it is decreasing. Equivalently, look at the sign of the difference quotient:
f(x) is increasing on R and a is a real number. How do f(a2+1) and f(2a) compare?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
An increasing function carries the order of the inputs straight over to the values, so just compare a2+1 with 2a:
a2+1−2a=(a−1)2≥0. When a=1 the two are equal, and so are the values; otherwise a2+1>2a. So f(a2+1)≥f(2a), with equality when a=1. Choosing “>” misses exactly this case.
Why: subtract, factor, decide the sign
Example: prove that f(x) = x³ − 3x is decreasing on [−1, 1]
Step 1 / 5
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Take any two points
Take any x1<x2 in [−1,1]; we must show f(x1)>f(x2). The pair in the figure is −0.5 and 0.5: high on the left, low on the right. But one pair proves nothing; we need every pair.
−1≤x1<x2≤1
Subtract and pull out x₂ − x₁
Decide the sign of the second factor
Conclude
Leave the interval
A few useful facts about monotone functions: the sum of two increasing functions is increasing; if f is increasing then −f is decreasing; and a composite function y=f(g(x)) follows the rule “same means increasing, different means decreasing”: if the inner and outer functions are both increasing or both decreasing, the composite is increasing; if one increases and the other decreases, it is decreasing.
Remove one condition
Break it
Counterexample lab: remove one condition
Conclusion: f(x)=x1 is decreasing← still true?
On the piece [0.5,3] of (0,+∞), every pair of points is “falling”: x21−x11=x1x2x1−x2<0, because x1x2>0.
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Widen the interval to [−3,3], which straddles 0. Put x1 at −1 and x2 at 1:
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This pair is “rising”. The fact x1x2>0 used in the proof no longer holds, and the difference quotient −x1x21 has become positive. x1 is decreasing on (−∞,0) and on (0,+∞)separately, but it is not decreasing on its whole domain.
Practice
Exercise 1Basic
The function f(x)=x2−2ax+3 is increasing on [2,+∞). Find the largest possible value of a.
Hint · direction
A parabola opening upward is increasing on which side of its axis?
Hint · method
The axis is x=a, and the function is increasing on [a,+∞). The whole of [2,+∞) must lie inside it.
Hint · first step
We need [2,+∞)⊆[a,+∞), that is, a≤2.
Full solution
The largest value of a is 2. (With a=2 the axis sits exactly at the left end of the interval, and the function is still increasing.)
Exercise 2Basic
f(x) is decreasing on R. What is the largest integer x with f(2x−1)>f(3)?
Hint · direction
For a decreasing function, a larger value means what about the input?
Hint · method
Turn f(2x−1)>f(3) into an inequality between the inputs. Watch the direction.
Hint · first step
2x−1<3.
Full solution
x<2, so the largest integer is 1.
Exercise 3Advanced
f(x)=x5+x3+x. How many integers x satisfy f(x2−2)+f(x)<0?
Hint · direction
What symmetry does f have? What monotonicity?
Hint · method
f is odd, and it is a sum of three increasing functions, so it is increasing. Move f(x) to the right-hand side, where it becomes f(−x).
Hint · first step
f(x2−2)<−f(x)=f(−x), so x2−2<−x.
Full solution
x2+x−2<0, (x+2)(x−1)<0, −2<x<1. The integers are −1 and 0: 2 of them.
Exercise 4Challenge
f(x) is increasing on (0,+∞), f(xy)=f(x)+f(y) for all positive x,y, and f(2)=1. Among the x satisfying f(x)+f(x−3)≤2, what is the largest?
Hint · direction
Combine the left side into a single f(⋯), and write the 2 on the right as f(⋯) too.
Hint · method
f(x)+f(x−3)=f(x(x−3)) and 2=f(2)+f(2)=f(4). Then use monotonicity to remove f, and do not forget the domain.
Hint · first step
We need x>0, x−3>0 and x(x−3)≤4.
Full solution
x2−3x−4≤0 gives −1≤x≤4; intersecting with x>3 gives 3<x≤4. The largest is x=4.
Decreasing on two pieces, so decreasing on their union
✗
The wrong step: f(x)=x1 is decreasing on (−∞,0)∪(0,+∞).
What went wrong? Think first, then open
DiagnosisTake x1=−1<x2=1: f(−1)=−1<f(1)=1, so the right-hand point is higher. Monotonicity must hold for any two points of the interval, and a pair straddling 0 breaks it.
✓
The right step: Write “f(x) is decreasing on (−∞,0) and on (0,+∞)”, separating the intervals with “and” or a comma. Do not join them with ∪.
“Do not glue monotone intervals together carelessly”
Exhibit
“Proving” monotonicity with a few special values
✗
The wrong step: f(1)<f(2)<f(3), so f(x) is increasing on [1,3].
What went wrong? Think first, then open
DiagnosisThree points say nothing about what happens in between: f(x)=(x−1.2)2 has f(1)=0.04<f(2)=0.64<f(3)=3.24, yet it is decreasing on [1,1.2].
✓
The right step: A proof must work for anyx1<x2: form the difference f(x2)−f(x1), turn it into a product, and decide the sign of each factor. Special values can only disprove monotonicity.