MAT01 Mathematics for Management and Economics

Monotony

Determine intervals where

\(f(x)=-12x^2+8x-1\)

is strictly monotonic increasing or strictly monotonic decreasing.

Preliminary remarks

Here you must find values x for which f is increasing or decreasing, respectively.

To do so, you need to link the idea of monotonic behaviour of a function to the 1st order derivative of that function (see theorem III.2.3.1 in block III learning sequence 2 of our script).

Formal approach

(i) Determine sections (intervals) where f is strictly monotonic increasing

All we need to do is determine the 1st order derivative f'(x) of f(x) and find values x for which f'(x) is positive.

The derivative is \(f'(x)=-24x+8 \),

and our «positive value condition» \(f'(x)=-24x+8 >0\) is equivalent to

\(8 > 24x\)

Division through the (positive) number 24 yields the equivalent condition

\(\frac {1}{3} > x \quad \) (since we divided by a positive number, the sense of the inequation remains unaltered).

Our conclusion is: for values x< 1/3, f(x) is strictly monotonic increasing
(don't get this confused : f'(x) > 0 means that f, not f', is increasing !).

(ii) Determine sections (intervals) where f is strictly monotonic decreasing

This time, f'(x) needs to be negative.

Of course, the calculation is the same as above. We need \(f'(x)=-24x+8<0 \), which, by the reasons explained above, is equivalent to \(8<24x\), which in turn is just the same as \(1/3 < x\).

Our conclusion: f(x) is strictly monotonic decreasing for all x> 1/3.

Please note that we now know where the graph of f(x) increases and where it decreases when travelling to the right along the x-axis. And we know this without even looking at the graph of f(x). Of course it is reassuring to actually ckeck our results. To do so, look at the diagram.

The trick we used was linking the monotony of a function to its own 1st order derivative: f is strictly monotonic increasing if f'(x) > 0, and strictly monotonic decreasing if f'(x) < 0.

Here, again, a diagram illustrates our findings: whenever the graph of f' is below the x-axis. i.e. whenever f'(x) < 0, f itself is decreasing, and vice versa for positive sections of the graph of f'.

A different approach

If inequations make you feel uncomfortable, you might take a different approach:

First, determine roots of f'(x): \(f'(x)=-24x+8 = 0 \quad \Rightarrow \quad x=1/3\)

Then, pick an x-value to the left of that root, like, for example, x = 0.

Determine f'(x) for that value: \(f'(0)=-24 \cdot 0+8 = 8 > 0\)

Thus, f is strictly monotonic increasing for any x< 1/3.

Likewise, since \(f'(1)=-24 \cdot 1+8 = -16 < 0\), f is strictly monotonic decreasing for any value x > 1/3.

But: this approach does not work in every case. See the following two examples to understand why.

Example 1:

The function \(f(x)=2x\) has \(f'(x)=2\) for every x.
Since 2 > 0, there is no root of \(f'(x)\) at all, and our conclusion must be that f is strictly monotonic increasing on its entire domain, i.e. for all real numbers x.

In this first example, it was quite easy to determine the monotony of the function. In the following, things are more complicated:

Example 2:

The function \(f(x)=\frac {1}{3}x^3-4x+1\) is strictly monotonic increasing for x < -2 and for x > 2, but strictly monotonic decreasing for -2 < x < 2.

Our "different approach" would have to look as follows:

\(f'(x)=x^2-4=0 \quad \Rightarrow \quad {x_1}=-2 \;\) and \(\;{x_2}=2\).

Since we have two different roots, the x-axis has to be divided into three intervals: one to the left of x₁, one to the right of x₂, and one between these values.

This means that we consecutively look at values x < -2, -2 < x < 2, and x > 2.

We compute \(f'(x)\) for any randomly chosen x of each interval:

\(f'(-3)=(-3)^2-4=5 \gt 0\), so f is strictly monotonic increasing for x < -2

\(f'(0)=0^2-4=-4\lt 0\), so f is strictly monotonic decreasing for -2< x <2

\(f'(3)=3^2-4=5 \gt 0\), so f is strictly monotonic increasing for x > 2

Please note that the idea behind our "different approach" can still be applied, but the steps to be taken need to be modified.

\(f'(x)=-24x+8\)

Strictly monotonic increasing:

\(f'(x)=-24x+8 \gt 0\quad \Rightarrow \quad 1/3 \gt x \quad \Rightarrow \quad \) f(x) is strictly monotonic increasing for \(x \lt 1/3\)

Strictly monotonic decreasing:

\(f'(x)=-24x+8 \lt 0\quad \Rightarrow \quad 1/3 \lt x \quad \Rightarrow \quad \) f(x) is strictly monotonic increasing for \(x \gt 1/3\)