Math Practice Worksheets – Functions

We hope these worksheets help you practice and assess your understanding of functions, one of the main topics in pre-calculus. We think that a facility with the questions here is evidence of a solid foundation and stands you in good stead for your study of calculus and beyond. 

Please write to us at info@wordpress-526411-2682118.cloudwaysapps.com/library if you have any questions, suggestions or topic requests. We plan to regularly expand existing worksheets and add worksheets on new topics.

 

Functions (Composition and Inverse):

Composition:

 

  1. Decompose \(f(x) = 2*((x+2)^2 +1)^2\) into a composition of scaling, shifting and squaring functions.

  2. Write \(f(x) = x^2 + x\) as a composition of the squaring function \(T(x) = x^2\) and shifting functions \(T_{a}(x) = x + a\). (a is some real number.)

  3. Given \(f(x) = x-2\) and \(g(f(x)) = 7x + 1\), what are \(g(x)\) and \(f(g(x))\)?

  4. Given \(f(x) = x + 2\) and \(g(f(x)) = x^2 + x\), what is \(f(g(x))\)?

  5. Given \(f(x) = ax + b\), \(g(x) = cx +d\) and that \(f(g(x)) = g(f(x))\), what relation do a, b, c, d obey?

  6. The graph of the function \(f(x)\) is transformed as follows (in this order):

    • Scaled horizontally by 2.
    • Shifted to the right by 3.
    • Reflected about the y-axis.
    • Reflected about the x-axis.
    • Shifted vertically by 5.
    • Scaled vertically by 2.

       

      Each of these transformations can be implemented as the action of a function (on the input or output of the function being transformed). What are these transforming functions? Write down the transformed \(f(x)\) as a composition of these transforming functions with \(f(x)\).

Inverse:

 

For each function below, does the inverse exist, and if so, what is it? 

  1. \(f(x) = \frac{1}{x+1}\)  (The domain is all \(\infty < x < \infty\) except \(x=1\))

  2. \(f(x) = |\sqrt{x}| \)  (The domain is \( 0 \leq x < \infty\))

  3. \(f(x) = x^2 + 4x +1\)  (Domain: \(-\infty < x < \infty\))

  4. \(f(x) = (2x+5)^3 +1\)  (Domain: \(-\infty < x < \infty\))

Composition:

 
  1. Consider the following functions: \(p(x) = x+2\),  \(q(x) = x^2\),  \(r(x) = x+ 1\)  and  \(s(x) = 2*x\). Working from the inside out, \(f(x) = s(q(r(q(p(x)))))\).

  2. Since we’re interested in squares and shifts, complete the square to write \(x^2 + x = \big(x+\frac{1}{2}\big)^2 – \frac{1}{4}\). Thus \(f(x) = T_{-\frac{1}{4}}(T(T_{\frac{1}{2}}(x)))\), where \(T(x) = x^2\),  \(T_{\frac{1}{2}}(x) = x + \frac{1}{2}\) and \(T_{-\frac{1}{4}}(f) = f – \frac{1}{4} \). 

  3. We know that \(f(x) = x-2\). Since, \(g(f(x)) = 7(x-2) + 15\), we can read off the “outer action” of g as: \(g(x) = 7x + 15\). Then \(f(g(x)) = (7x+15)-2 = 7x +13\).

  4. Let \(g(x) = ax^2 +bx +c\). Then \(g(f(x)) = a(x+2)^2 + b(x+2) + c\). Matching coefficients we get, \(a=1, b = -3\) and \(c=2\). Thus, \(f(g(x)) = (x^2 -3x +2) +2 = x^2 -3x +4)\).

  5. \(f(g(x) = a(cx +d) + b = (ac)x + (ad+b) = g(f(x)) = c(ax+b) +d = (ca)x + (cb +d)\). Matching coefficients of each power of \(x\) gives the condition: \(ad +b = cb + d\). Rewriting, \(d(a-1) = c(b-1)\).

  6. Lets describe the effect of the transformations on a general function \(f(x)\). Call the transformed function \(Tf(x)\). Here are the functions which implement the various transformations:

    • \(T_1(x) = \frac{1}{2}x\) transforms a graph by stretching it horizontally by a factor of 2.
    • \(T_2(x) = x – 3\) shifts a graph horizontally to the right by 3.
    • \(T_3(f) = -f\) reflects about the x-axis.
    • \(T_4(x) = -x\) reflects about the y-axis.
    • \(T_5(f) = f +5\) shifts vertically by a factor of 5.
    • \(T_6(f) = 2f\) scales vertically by a factor of 2.

      Thus, \(Tf(x) ~=~ T_6(T_5(T_3(f(T_1(T_3(T_4(x))))))\). We recommend carefully building up this expression yourself, there are a lot of brackets!

Inverse:

 

  1. Yes, \(f(x)\) is one to one. (Note that the graph of \(f(x)\) is just the hyperbola \(y = \frac{1}{x}\) shifted to the left by 1). Manipulating the relationship between f and x gives, \(x = 1\frac{1}{f}-1\). Thus, \(f^{-1}(x) = \frac{1}{x} -1\).

  2. Yes, \(f(x)\) is one to one since the modulus retains only the “positive square root”. Squaring both sides, \(f^2 = x\). Thus the inverse function is \(f^{-1}(x) = x^2\).

  3. No. \(f(x)\) is not a one to one function since there exist values of f for which two values of x will satisfy \(f = x^2 + 4x +1\). Since, \(f(x) = (x+2)^2 -3)\), it is a composition of the functions \(f(x) = x +2\), \(f(x) = x^2\) and \(f(x) = x -3\). However \(f(x) = x^2\) is not one to one. Thus \(f(x)\) fails to be one to one and hence is not invertible.

  4. Yes. First note that the cubic function \(f(x) = x^3 \) is one to one for real inputs (unlike the \(y=x^2\) the graph does not curve back up). Next, the function \(f(x) = 2x +5\) (graphically, a straight line) is also one to one, as is the shift function \(f(x) = x+1\).  \(f(x) = (2x+5)^3 +1\) is a composition of these three functions and hence is also one to one and thus invertible. Manipulating the relationship between f and x gives, \(x = \frac{\sqrt[3]{f – 1}-5}{2}\). Thus, \(f^{-1}(x) = \frac{\sqrt[3]{x-1}-5}{2}\).

This accessible lecture from MIT discusses inverse functions and focuses on the example of the exponential and logarithmic functions.

Periodic Functions:

  1. What is the period of the function \(f(x) = tan(3x)\)?

  2. Is \(f(x) = sec(2x +3)\) a periodic function? If so what is its period?

  3. Is \(f(x) = sin(2x) + sin(3x)\) a periodic function? If so what is its period?

  4. Is \(f(x) = sin(2x) + tan(3x)\) a periodic function? If so what is its period?

  5. Consider \(f(x) = sin(ax) + sin(bx)\), where a and b are integers. What is its period?

  6. Is \(f(x) = sin(x) + sin(\pi x)\) a periodic function?

  1. The period of \(f(x) = tan(x)\) is \(\pi\). Note that \(sin(x+\pi) = -sin(x)\) and \(cos(x+\pi) = -cos(x)\). Thus, \(tan(x+\pi) = \frac{cos(x+\pi)}{sin(x+\pi)} = tan(x)\). Finally the period of \(f(x) = tan(3x)\) is one third of the period of \(tan(x)\), that is \(\frac{\pi}{3}\).

  2. Yes. Note that \(sec(x) = \frac{1}{cos(x)}\). So the period of \(sec(x)\) is also \(2\pi\) and the period of \(sec(2x)\) is \(\pi\). Since \(f(x) = sec(2x + 3)\) is obtained from \(sec(x)\) simply by shifting its graph by 3 in the -x direction, its period is also \(pi\).

  3. Yes. Note that \(f(x+2\pi)) = sin(2x + 4\pi) + sin(3x+6\pi) = sin(2x) + sin(3x)\). \(sin(2x)\) has a period of \(pi\) while \(sin(3x)\) has a period of \(\frac{2\pi}{3}\). Since \(2 \times\pi = 3 \times \frac{2\pi}{3}\ = 2\pi\), both functions repeat themselves after an interval of \(2\pi\).

  4. Yes. The period of \(sin(2x)\) is is \(\pi\) while that of (\tan(3x)\) is \(\frac{\pi}{3}\). Since \(3\times\frac{\pi}{3} = \pi\), the period of \(f(x) = sin(2x) + tan(3x)\) is \(pi\).

  5.  Yes. The period of \(sin(ax)\) is \(\frac{2\pi}{a}\), while the period of \(sin(bx)\) is \(\frac{2\pi}{b}\). So we have to find the smallest two integers, m and n such that \(m\times \frac{2\pi}{a} = n\times \frac{2\pi}{b}\): this gives interval which equals some number of whole “cycles” of both functions. So both functions and hence their sum repeat over this interval. Our equation may be rewritten as, \(mb = na\). Thus the smallest m, n are such that \(mb\), \(na\) equal LCM(a,b) and \(m = \frac{LCM(a,b)}{a}\), \(n = \frac{LCM(a,b)}{b}\). Thus the period is, \(2\pi\frac{LCM(a,b)}{ab}\).

  6. No. Let’s ask a more general question: when is \(f(x) = sin(ax) + sin(bx)\) a periodic function? (a and b are real numbers). If there exist integers m, n such that \(m\times \frac{2\pi}{a} = n \times\frac{2\pi}{b} ~\rightarrow~ \frac{b}{a} = \frac{n}{m}\), that is if the ratio of a to b is rational, then \(sin(ax)\) and \(sin(bx)\) have a common period and thus \(f(x)\) is a periodic function. However, if this is not the case, then \(f(x)\) is not a periodic function. Let’s prove this (the only proof I know uses the derivative):

    The strategy is to arrive at a contradiction. Note, that \(\frac{d^2}{dx^2}~sin(ax) ~=~ -a^2~sin(ax)\). Thus we have (EQN 1) \[f(x) + \frac{1}{a^2}\frac{d^2}{dx^2}f(x) ~=~ sin(bx) -\frac{b^2}{a^2} ~sin(bx).\] Similarly (EQN 2), \[f(x) + \frac{1}{b^2}\frac{d^2}{dx^2}f(x) ~=~ sin(ax) -\frac{a^2}{b^2} ~sin(ax).\] Now if it were true that \(f(x+P) = f(x)\) for some \(P\), then it must be that \(P = n \times \frac{2\pi}{b}\) for some integer \(n\), since the RHS of EQN 1 is periodic only over multiples of \(\frac{2\pi}{b}\), while the LHS has a period \(P\). Similarly, all periods of the RHS of EQN 2 are mutiples of \(\frac{2\pi}{a}\) and so it must also be that \(P = m \times \frac{2\pi}{a}\) for some integer \(m\). But this implies, \(m\frac{2\pi}{a} ~=~ n\frac{2\pi}{b} ~\rightarrow~ \frac{b}{a} = \frac{n}{m}\). Since we had assumed that this was not the case, we have a contradiction and so \(f(x)\) cannot be a periodic function.

A remarkable and extremely useful fact is that any periodic function can be written as a sum of sines and cosines (given that it satisfies some conditions pertaining to any discontinuities it has). Such a decomposition of a periodic function is called its “Fourier series”. This video by Dr. Trefor Bazett has a clear introduction to the idea of Fourier series.