Mathematics · Glossary

What is Increasing, decreasing?

Also known as: increasing function · decreasing function

Definition 3.8 High School Mathematics · Chapter 3 — Functions

Let ff be defined on an interval II.

  • ff is increasing on II when for all u,vIu, v \in I, if u<vu < v then f(u)f(v)f(u) \leq f(v): the outputs grow with the inputs, and the graph climbs from left to right.
  • ff is decreasing on II when u<vu < v implies f(u)f(v)f(u) \geq f(v): the graph falls from left to right.

With strict inequalities f(u)<f(v)f(u) < f(v) (resp. >>) we say strictly increasing (resp. decreasing).

Examples

Example 3.10

Consider the function graphed below on [2,4]\intcc{-2}{4}.

A function defined on [-2, 4], increasing then decreasing.
A function defined on [2,4]\intcc{-2}{4}, increasing then decreasing.

Reading the graph: ff increases from f(2)=1.5f(-2) = -1.5 up to its maximum f(1)=3f(1) = 3, then decreases down to f(4)=1.5f(4) = -1.5. Its variation table is

xx2-21144
ff1.5-1.5\nearrow33\searrow1.5-1.5

Example 3.11 (Proving a variation)

Show that f(x)=3x+1f(x) = 3x + 1 is strictly increasing on R\R. Take any u<vu < v and compare the images:

f(v)f(u)=(3v+1)(3u+1)=3(vu)>0,f(v) - f(u) = (3v + 1) - (3u + 1) = 3(v - u) > 0,

because vu>0v - u > 0. So f(u)<f(v)f(u) < f(v): the function is strictly increasing. The same computation with a negative slope, e.g. g(x)=2x+5g(x) = -2x + 5, gives g(v)g(u)=2(vu)<0g(v) - g(u) = -2(v-u) < 0: strictly decreasing.

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Definition 11.4 High School Mathematics · Chapter 11 — Functions and Variations

Let ff be defined on an interval II. The function ff is increasing on II if it preserves order:

for all u,vI:u<v    f(u)f(v),\text{for all } u, v \in I: \quad u < v \implies f(u) \leq f(v),

and decreasing if it reverses order (u<v    f(u)f(v)u < v \implies f(u) \geq f(v)). With strict inequalities on the right, ff is strictly increasing (resp. decreasing). A function that is increasing or decreasing on II is monotonic on II.

Examples

Example 11.10

Study f(x)=x2+1f(x) = \sqrt{x^2 + 1} on R\R. The inner function u(x)=x2+1u(x) = x^2 + 1 is positive, decreasing on (,0]\intoc{-\infty}{0} and increasing on [0,+)\intco{0}{+\infty} (a shifted square). By point 3, ff has the same variations: decreasing then increasing, with minimum f(0)=1f(0) = 1.

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