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Differentiation I

Increasing and decreasing functions

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Variable acceleration in one dimension

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Functions of time: Displacement, velocity, acceleration

Using differentiation to find velocity and acceleration

Maxima and minima problems

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Differentiation II

Differentiating sin x and cos x

Differentiating exponentials and logarithmic functions

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Differentiating trigonometric functions

Parametric differentiation

Implicit differentiation

Second derivatives: Concave and convex functions

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Integration I

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Differentiation I

Finding the gradient of a curve

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Increasing and decreasing functions

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Summary

Increasing and decreasing functions

In a nutshell

In a given interval, functions could only be getting bigger (increasing) or smaller (decreasing). This can easily be checked by finding the gradient of the function in the interval since the gradient gives a measure of how a function is increasing or decreasing.

Interval notation

Intervals can be noted with either curved brackets or square brackets. Curved brackets do not include the values stated as the boundaries, whereas square brackets do. For example, the interval of $(4,8)$ represents any number $x$ such that $4<x<8$ . On the other hand, $[4,8]$ means any value $x$ such that $4\leq x\leq8$ .

Increasing functions

The function $f(x)$ is increasing in the interval $[a,b]$ if its derivative with respect to $x$ is greater than or equal to zero for values of $x$ in the interval. In other terms, $f(x)$ is increasing on $[a,b]$ if $f'(x)\geq0$ for $a\leq x\leq b$ .

You can mix the brackets if on one side you want to include the boundary value, but on the other you don't. For instance, the interval

$-3<x\leq5$

is denoted with

$(-3,5]$

If you want to use all positive or all negative values, you will want to use infinity. As boundaries, these require curved brackets. All real numbers are represented by the notation:

$(-\infty,\infty)$

Example 1

Show that the function $f(x)=4x^3+2x$ is increasing in the interval $[5,10]$ .

First differentiate the function with respect to $x$ :

$f'(x)=12x^2+2$ 

This is in fact always positive, since $x^2$ is always greater or equal to zero. Thus not only is this function increasing on the given interval, it is increasing for all values of $x$ .

Decreasing functions

The function $f(x)$ is decreasing in the interval $[a,b]$ if its derivative with respect to $x$ is less than or equal to zero for values of $x$ in the interval. In other terms, $f(x)$ is decreasing on $[a,b]$ if $f'(x)\leq0$ for $a\leq x\leq b$ .

Example 2

Show that the function $g(x)=-4x^2$ is decreasing in the interval $[0,\infty)$ .

First differentiate the function:

$g'(x)=-8x$ 

Thus in the interval of all non-negative real values, this derivative is always negative. So the function is decreasing in this interval.

Finding the interval

Sometimes you will not be given the interval and will have to find it.

Example 3

Find the interval(s) in which the function $h(x)=2x^3-15x^2-36x+4$ is increasing.

First differentiate:

$h'(x)=6x^2-30x-36$ 

Thus you are looking for an interval such that $h'(x)\geq0$ .

$6x^2-30x-36\geq0\\\space\\x^2-5x-6\geq0\\\space\\(x+1)(x-6)\geq0$ 

By sketching this curve, you will find that

$x\leq-1$ and $6\leq x$

With interval notation, you can express this as

$(-\infty,-1]$ and $[6,\infty)$ 

Or more formally:

$\underline{x\in(-\infty,-1] \cup[6,\infty)}$

Note: The $\in$ means that " $x$ is in" the interval. The $\cup$ is used because $x$ is in either of these intervals. It cannot simultaneously be in both intervals.

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