HOW TO DETERMINE IF A FUNCTION IS MONOTONIC

Definition :

A function that is completely increasing or completely decreasing on the given interval is called monotonic on the given interval.

Usually, by looking at the graph of the function one can say whether the function is increasing or decreasing or neither.

The graph of an increasing function does not fall as we go from left to right while the graph of a decreasing function does not rise as we go from left to right.

But if we are not given the graph, how do we decide whether a given function is monotonic or not ?

Test for monotonic functions :

Let I be an open interval. Let f : I → R be differentiable. Then

(i) The function f is increasing if and only if f ′(x) ≥ 0 for all x in I.

(ii) The function f is decreasing if and only if f ′(x) ≤ 0 for all x in I.

Note :

If a function changes its signs at different points of a region (interval) then the function is not monotonic in that region. So to prove the non- monotonicity of a function, it is enough to prove that f ′ has different signs at different points.

Example 1 :

Check whether the function

y = sin x + cos 2x

is monotonic on the interval [0, Π/4]

Solution :

f(x)  =  sin x + cos 2x

f'(x)  =  cos x - sin 2x [2 (1)]

f'(x)  =  cos x - 2 sin 2x

By applying  x = 0, we get

f'(0)  =  cos 0 - 2 sin 2(0)

f'(0)  =  1-2 (0)

f'(0)  =  1 - 0

f'(0)  =  1 > 0 

By applying x = Π/4, we get

f'(Π/4)  =  cos Π/4 - 2 sin 2(Π/4)

f'(Π/4)  =  1/√2 - 2 sin (Π/2)

f'(Π/4)  =  1/√2 - 2 (1)

f'(Π/4)  =  1/√2 - 2

f'(Π/4)  =  0.707 - 2

f'(Π/4)  =  -1.292 < 0

Thus f′ is of different signs at 0 and π/4. So, the given function is not monotonic function on the interval [0, Π/4].

Example 2 :

Check whether the function

y = x sin x

is monotonic on the interval [Π/2, Π]

Solution :

f(x)  =  x sin x

f'(x)  =  x (cos x) + sin x (1)

f'(x)  =   x (cos x) + sin x 

By applying x = Π/2

f'(0)  =  Π/2 (cos Π/2) + sin Π/2

f'(0)  =  Π/2 (0) + 1

f'(0)  =  1 > 0

By applying x = Π

f'(Π)  =  Π (cos Π) + sin Π

f'(0)  =  Π (-1)  + 0

f'(0)  =   - Π < 0

So, the function is not monotonic function.

Example 3 :

Discuss monotonicity of the function

(i)  ex for all real numbers.

(ii) log x on (0, ∞)

Solution :

(i)  f(x)  =  ex 

Let f(x)  =  ex

f'(x)  =  ex

  • If x > 0, then f′(x) > 0. The function is strictly increasing for all positive values of x.
  • If x < 0 then f′(x) > 0. The function is strictly increasing for all negative values of x.
  • If x  =  0, then f'(x)  =  1,we get positive value for f'(x).

So, the function monotonic on real numbers.

(ii)  f(x)  =  log x on (0, ∞)

Let f(x)  =  log x

f'(x)  =  1/x

If x > 0, then f′(x) > 0. The function is strictly increasing for all positive values of x.

So, the function monotonic on (0, ∞).

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