NZ Level 8 (NZC) Level 3 (NCEA) [In development]

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.

The quadratic function $f$`f` defined on the real numbers given by $f(x)=x^2+3x-7$`f`(`x`)=`x`2+3`x`−7 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`)=2`x`+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.

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

Consider the function given by the rule $g(x)=\frac{1}{2}\left(e^x-e^{-x}\right)$`g`(`x`)=12(`e``x`−`e`−`x`) 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(`e``x`+`e`−`x`)

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(`e``x`+`e`−`x`) 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.

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

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

Stationary points occur where $f'(x)=0$`f`′(`x`)=0. So, we put $3x^2-4x-2=0$3`x`2−4`x`−2=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±√16−4×3×(−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`=2−√103 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)2−4×(−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`=2−√103 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)2−4×(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.

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

Apply differentiation methods in solving problems