NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Indentify increasing/decreasing intervals and stationary points
Lesson

We find the derivative of a function when we want information about how its gradient varies over its domain. A function may be increasing on certain intervals, or decreasing, and there may be points, called stationary points, where the gradient is zero.

A function is said to be decreasing everywhere in an open interval if its gradient function (the derivative) is negative at all points in the interval.

A function is said to be increasing on an open interval if its gradient function is positive at all points in the interval.

We can also speak of a function being non-increasing or non-decreasing if we wish to include the possibility of the gradient being zero at one or more points within the interval. If a function is non-increasing or non-decreasing over the whole of its domain, it is said to be a monotonic function - monotonically decreasing or monotonically increasing.

 

Example 1

The quadratic function $f$f defined on the real numbers given by $f(x)=x^2+3x-7$f(x)=x2+3x7 is increasing over part of its domain and non-increasing elsewhere.  

We wish to find precisely where the different kinds of gradient occur. We differentiate to obtain

$f'(x)=2x+3$f(x)=2x+3

Since $f'(x)>0$f(x)>0 when $x>-\frac{3}{2}$x>32, and nowhere else, we can conclude that $f$f is increasing on the interval $(-\frac{3}{2},\infty)$(32,). It must be non-increasing on the interval $(-\infty,-\frac{3}{2}]$(,32].

Notice that this latter interval is closed on the right since it includes the endpoint $x=-\frac{3}{2}$x=32. With a little more work, we can see that $f'(x)=0$f(x)=0 when $x=-\frac{3}{2}$x=32, and nowhere else. This is the only stationary point belonging to this function.

 

Example 2

In this example, we need to know that the derivative of the exponential function $e^x$ex where $e$e is the special exponential base equal to approximately $2.7$2.7, is its own derivative. That is $\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x$ddxex=ex.

Consider the function given by the rule $g(x)=\frac{1}{2}\left(e^x-e^{-x}\right)$g(x)=12(exex) where $x$x can be any real number. Does it have a stationary point? Describe how its gradient varies.

As before, we differentiate. The gradient function is

$g'(x)=\frac{1}{2}\left(e^x+e^{-x}\right)$g(x)=12(ex+ex)

Next, we ask, for what values of $x$x is $g'(x)=0$g(x)=0? Clearly $\frac{1}{2}\left(e^x+e^{-x}\right)$12(ex+ex) is never $0$0 because the exponential expressions are always positive. So, $g$g cannot have a stationary point.

Moreover, the observation that $g'(x)$g(x) is always positive shows that $g$g is an increasing function over the whole of its domain. In fact, its gradient is never less than $1$1.

 

Example 3

Find any stationary points for the function $f:R\rightarrow R,f(x)=x^3-2x^2-2x-1$f:RR,f(x)=x32x22x1 and intervals over which the function is decreasing.

By differentiation, $f'(x)=3x^2-4x-2$f(x)=3x24x2.

Stationary points occur where $f'(x)=0$f(x)=0. So, we put $3x^2-4x-2=0$3x24x2=0 and solve.

By the quadratic formula, we see that the solutions are $x=\frac{4\pm\sqrt{16-4\times3\times(-2)}}{6}=\frac{2\pm\sqrt{10}}{3}$x=4±164×3×(2)6=2±103

That is, the stationary points are at approximately $x=-0.387$x=0.387 and $x=1.721$x=1.721.

These are the only stationary points, so the function must be either increasing or decreasing between these points. If we pick a point between the stationary points, say $x=0$x=0, we find $f'(0)=-2$f(0)=2 and so, the function must be decreasing on the interval between the stationary points.

To the left of $x=\frac{2-\sqrt{10}}{3}$x=2103 at say, $x=-0.4$x=0.4, we have $f'(x)=3\times(-0.4)^2-4\times(-0.4)-2=0.08$f(x)=3×(0.4)24×(0.4)2=0.08, $f'(x)$f(x) is positive and so, $f$f is increasing.

The stationary point $x=\frac{2-\sqrt{10}}{3}$x=2103 must be the location of a local maximum.

To the right of $x=\frac{2+\sqrt{10}}{3}$x=2+103 at say, $x=1.8$x=1.8, we have $f'(x)=3\times(1.8)^2-4\times(1.8)-2=0.52$f(x)=3×(1.8)24×(1.8)2=0.52$f'(x)$f(x) is again positive and so, $f$f is increasing in this region.

The stationary point $x=\frac{2+\sqrt{10}}{3}$x=2+103 must be the location of a local minimum since the function is decreasing to the left of this point but increasing to the right of it.

 

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91578

Apply differentiation methods in solving problems

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