Title: Create Examples
Course: Demonstrating STACK (v4.4.7)

Give an example of an even function.

\(f(x) = \)
Feedback: Remember, a function \(f(x)\) is said to be even if \[ f(x) = f(-x) \] for all values of \(x\). Hence, there are many examples which satisfy this definition, \(\cos(x)\), \(3*x^4\), and even \(f(x)=c\) for any constant.

Find a function of \(f(x)\) which satisfies \[ f(a+b)=f(a)+f(b)\] for all values of \(a\) and \(b\).

\(f(x) = \)
Feedback: One way to approach questions like this is to start with a general polynomial and try to fit coefficients. For example, starting with \(p(x)=ax^2+bx+c\) we would need \[ a(x_1+x_2)^2+b(x_1+x_2)+c = x_1^2+bx_1+c + x_2^2+bx_2+c\] for all values of \(x_1\) and \(x_2\). Take \(x_1=x_2=0\) means that \(c=0\). If \(x_1\) and \(x_2\) are not zero we also need \(a=0\), but any value of \(b\) will be fine. Hence, any straight line \(p(x)=bx\) is an example.

Give an example of a function \(f(x)\) with a stationary point at \(x=[value]\) and which is continuous but not differentiable at \(x=0\).

\(f(x)=\)
Feedback: In this question we can use the absolute value function to "build" and example which works. We want a function \(f(x)\) with a stationary point at \(x=[value]\) and which is continuous but not differentiable at \(x=0\). So, start with one of these properties, e.g. a continuous function with a stationary point at \(x=[value]\). That could be \(f(x)=(x-[value])^2\). Now, this only some of the properties we need. We want to modify this to not be differentiable at \(x=0\). In this form it is not so easy, so what about \(f(x)=x(x-2\times[value])\)? This has a stationary point at \(x=[value]\) but is easier to modify at \(x=0\) by using \[ f(x)=|x|(x-[value]).\] There are many other examples.

Give an example of an odd function.
Feedback: Remember, a function \(f(x)\) is said to be odd if \[ f(x) = -f(-x) \] for all values of \(x\). Hence, there are many examples which satisfy this definition, \(\sin(x)\), \(4*x^5\), and even \(f(x)=0\).

Give an example of a quadratic with roots at \(x=[value]\) and \(x=[value]\).

\(f(x) = \)
Feedback: A root of a polynomial \(p(x)\) is a point at which \(p(x)=0\). So, if we want a root at \(x=a\) the polynomial will have a factor of \((x-a)\). A quadratic with roots at \(x=[value]\) and \(x=[value]\) will have factors \((x-[value])\) and \((x-[value])\). Multiplying these together gives \[ p(x) = (x-[value])(x-[value]) \] which is already a quadratic.

Find a quadratic, \[ x^2 +bx +c, \] other than \(x^2\), which has roots equal to \(b\) and \(c\).

\(f(x)=\)
Feedback: Since \(b\) and \(c\) are the roots, we have \[ (x-b)(x-c) =0. \] Expanding this, and comparing coefficients, we have that \[ \left. \begin{array}{rcl} b+c = -b \\ bc = c \end{array} \right\} \] From the second of these equations, either \(b=1\) or \(c=0\). If \(c=0\), the first gives \(b=0\), resulting in \(x^2\). If \(b=1\), then \(c=-2\), giving only one other solution \[ (x+2)(x-1) = x^2 +x -2.\] (By Teiji Nakazawa, Mathematical Gazette, No. 358, pg 315, 1962)

Give an example of a function \(f(x)\) with a stationary point at \(x=[value]\).

\(f(x) = \)
Feedback: In this question you are asked to find a function with the property that there is a stationary point at \(x=[value]\). That is to say that the gradient here will be zero. There are two main ways to go about this - Take a known function and graph shift it to put the stationary point in the correct place. - Take a function with a zero in the correct place an integrate it. There are many different examples.