Root hunter
In this short problem, try to find the location of the roots of some unusual functions by finding where they change sign.
Problem
Take a look at the function in the graph below.
The graph is positive for $x = 5$ and negative for $x = 3$. This means that the graph must cut the $x$ axis somewhere between $3$ and $5$.
Although in this case the result is obvious (because we have the whole graph to look at!), we can also use this idea to show that more tricky functions also have roots.
Use this idea to show that these functions possess at least one solution $f(x) = 0$:
$$ f(x)=\frac{1}{x-2}+\frac{1}{x-3} $$ $$f(x)= x^x - 1.5 x$$ $$f(x)= x^{1000000}+{1000000}^x - 17$$ $$f(x)=\cos(\sin(\cos x)) - \sin(\cos(\sin x)) $$
Optional extension activity: Can you make a spreadsheet that helps you find the numerical values of the roots to, say, four decimal places?
Getting Started
When experimenting with numbers in equations it is often a good idea to try common numbers such as $0, 1, 2$ and $-1$. You might also try really big numbers such as a million or really small numbers such as $0.000001$.
Student Solutions
$\mathbf{f(x) =\frac{1}{x-2} + \frac{1}{x-3}}$
For the argument to be valid, we must ensure that our functions are continuous in the range in question. For example, $f(1.9) \approx -10.9$ and $f(2.1) \approx 8.9$, but there is no zero between these values since the function in not continuous at $x = 2$.
However $f(2.9) \approx -8.9$, and $f$ is continuous for $2 < x < 3$ so there is a root between $2.1$ and $2.9$.
$\mathbf{f(x) = x^x - 1.5x}$
$f(1) =1^1 - 1.5 = -0.5$
and $f(2) = 2^2 - 1.5\times 2 = 4 - 3 = 1$
so there is a root between $1$ and $2$.
$\mathbf{f(x) = x^{1000000} + 1000000^x - 17}$
$f(-1) = (-1)^{1000000} + 1000000^{-1} - 17 = 1 + 10^{-6} - 17 = -15.999999$
and $f(1) = 1^{1000000} + 1000000^1 - 17 = 1 + 1000000 - 17 = 999984$
so there is a root between $-1$ and $1$.
$\mathbf{f(x) = \cos{(\sin{(\cos{x})})} -\sin{(\cos{(\sin{x})})}}$
$f(0) = \cos{(\sin{1})} - \sin{(\cos{0})} = \cos{0.84147 \cdots} -\sin{1} = 0.66637\cdots - 0.84147\cdots = -0.1751\cdots$
and $f(\frac{\pi}{2}) = \cos{(\sin{0})} - \sin{(\cos{1})} = \cos{0} - \sin{0.54030\cdots} = 1 - 0.5144\cdots = 0.4856\cdots$
So there is a root between $0$ and $\frac{\pi}{2}$.
Teachers' Resources
Why do this problem?
This gives an interesting challenge in which students develop their understanding of functions and skills at approximation and estimation.
Possible approach
Students should be encouraged to try out function evaluation for various choices of numbers. Remember, that the key point is whether the function is positive or negative: we don't need to evaluate the exact values. Students should be encouraged to focus on whether the function is positive or negative, rather than computing the exact values.
Key Questions
What do we know about the values of a function either side of a solution?
Can you think of functions for which this sort of approach might not work?
Does this method tell us anything about the number of roots of an equation?
Possible extension
Note that at university this sort of idea is extended in courses on Analysis, and this result is called the Intermediate Value Theorem. It is actually a very useful and powerful mathematical idea.
Possible support
Suggest that students try key values of $0.5, 1, 1.5$ and so on. Give them calculators.Suggest that they tabulate the results of their calculations.