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Functions

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  1. Given \(f\left( x \right) = - {x^2} + 6x - 11\) and \(g\left( x \right) = \sqrt {4x - 3} \) find each of the following.
    1. \(f\left( 2 \right)\)
    2. \(g\left( 2 \right)\)
    3. \(f\left( { - 3} \right)\)
    4. \(g\left( {10} \right)\)
    1. \(f\left( t \right)\)
    2. \(f\left( {t - 3} \right)\)
    3. \(f\left( {x - 3} \right)\)
    4. \(f\left( {4x - 1} \right)\)
    Show Solution

    All throughout a calculus sequence you will be asked to deal with functions so make sure that you are familiar and comfortable with the notation and can evaluate functions.

    First recall that the \(f\left( x \right)\) in a function is nothing more than a fancy way of writing the \(y\)in an equation so

    \[f\left( x \right) = - {x^2} + 6x - 11\]

    is equivalent to writing

    \[y = - {x^2} + 6x - 11\]

    except the function notation form, while messier to write, is much more convenient for the types of problem you’ll be working in a Calculus class.

    In this problem we’re asked to evaluate some functions. So, in the first case \(f\left( 2 \right)\) is asking us to determine the value of \(y = - {x^2} + 6x - 11\) when \(x = 2\).

    The key to remembering how to evaluate functions is to remember that you whatever is in the parenthesis on the left is substituted in for all the \(x\)’s on the right side.

    So, here are the function evaluations.

    1. \(f\left( 2 \right) = - {\left( 2 \right)^2} + 6(2) - 11 = - 3\)
    2. \(g\left( 2 \right) = \sqrt {4(2) - 3} = \sqrt 5 \)
    3. \(f\left( { - 3} \right) = - {\left( { - 3} \right)^2} + 6( - 3) - 11 = - 38\)
    4. \(g\left( {10} \right) = \sqrt {4(10) - 3} = \sqrt {37} \)
    5. \(f\left( t \right) = - {t^2} + 6t - 11\)

      Remember that we substitute for the \(x\)’s WHATEVER is in the parenthesis on the left. Often this will be something other than a number. So, in this case we put \(t\)’s in for all the \(x\)’s on the left.

      This is the same as we did for (a) – (d) except we are now substituting in something other than a number. Evaluation works the same regardless of whether we are substituting a number or something more complicated.

    6. \(f\left( {t - 3} \right) = - {\left( {t - 3} \right)^2} + 6\left( {t - 3} \right) - 11 = - {t^2} + 12t - 38\)

      Often instead of evaluating functions at numbers or single letters we will have some fairly complex evaluations so make sure that you can do these kinds of evaluations.

    7. \(f\left( {x - 3} \right) = - {\left( {x - 3} \right)^2} + 6\left( {x - 3} \right) - 11 = - {x^2} + 12x - 38\)

      The only difference between this one and the previous one is that I changed the \(t\) to an \(x\). Other than that, there is absolutely no difference between the two!

      Do not let the fact that there are \(x\)’s in the parenthesis on the left get you worked up! Simply replace all the \(x\)’s in the formula on the right side with \(x - 3\). This one works exactly the same as (f).

    8. \(f\left( {4x - 1} \right) = - {\left( {4x - 1} \right)^2} + 6\left( {4x - 1} \right) - 11 = - 16{x^2} + 32x - 18\)

      Do not get excited by problems like (e) – (h). This type of problem works the same as (a) – (d) we just aren’t using numbers! Instead of substituting numbers you are substituting letters and/or other functions. So, if you can do (a) – (d) you can do these more complex function evaluations as well!

  2. Given \(f\left( x \right) = 10\) find each of the following.
    1. \(f\left( 7 \right)\)
    2. \(f\left( 0 \right)\)
    3. \(f\left( { - 14} \right)\)
    Show Solution

    This is one of the simplest functions in the world to evaluate, but for some reason seems to cause no end of difficulty for students. Recall from the previous problem how function evaluation works. We replace every \(x\)on the right side with what ever is in the parenthesis on the left. However, in this case since there are no \(x\)’s on the right side (this is probably what causes the problems) we simply get 10 out of each of the function evaluations. This kind of function is called a constant function. Just to be clear here are the function evaluations.

    \[f\left( 7 \right) = 10\hspace{0.5in}f\left( 0 \right) = 10\hspace{0.5in}f\left( { - 14} \right) = 10\]
  3. Given \(f\left( x \right) = 3{x^2} - x + 10\) and \(g\left( x \right) = 1 - 20x\) find each of the following.
    1. \(\left( {f - g} \right)\left( x \right)\)
    2. \(\left( {\frac{f}{g}} \right)\left( x \right)\)
    3. \(\left( {fg} \right)(x)\)
    1. \(\left( {f \circ g} \right)\left( 5 \right)\)
    2. \(\left( {f \circ g} \right)\left( x \right)\)
    3. \(\left( {g \circ f} \right)\left( x \right)\)
    Show Solution

    This problem makes sure you are familiar with notation commonly used with functions. The appropriate formulas are included in the answer to each part.

    1. \begin{align*}\left( {f - g} \right)\left( x \right) & = f\left( x \right) - g\left( x \right)\\ & = 3{x^2} - x + 10 - \left( {1 - 20x} \right)\\ & = 3{x^2} - x + 10 - 1 + 20x\\ & = 3{x^2} + 19x + 9\end{align*}
    2. \begin{align*}\left( {\frac{f}{g}} \right)\left( x \right) & = \frac{{f\left( x \right)}}{{g\left( x \right)}}\\ & = \frac{{3{x^2} - x + 10}}{{1 - 20x}}\end{align*}
    3. \begin{align*}\left( {fg} \right)\left( x \right) & = f\left( x \right)g\left( x \right)\\ & = \left( {3{x^2} - x + 10} \right)\left( {1 - 20x} \right)\\ & = - 60{x^3} + 23{x^2} - 201x + 10\end{align*}
    4. For this part (and the next two) remember that the little circle, \( \circ \), in this problem signifies that we are doing composition NOT multiplication!

      The basic formula for composition is

      \[\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\]

      In other words, you plug the second function listed into the first function listed then evaluate as appropriate.

      In this case we’ve got a number instead of an \(x\) but it works in exactly the same way.

      \[\begin{align*}\left( {f \circ g} \right)\left( 5 \right) & = f\left( {g\left( 5 \right)} \right)\\ & = f\left( { - 99} \right)\\ & = 29512\end{align*}\]
    5. Compare the results of this problem to (c)! Composition is NOT the same as multiplication so be careful to not confuse the two!

      \[\begin{align*}\left( {f \circ g} \right)\left( x \right) & = f\left( {g\left( x \right)} \right)\\ & = f\left( {1 - 20x} \right)\\ & = 3{\left( {1 - 20x} \right)^2} - \left( {1 - 20x} \right) + 10\\ & = 3\left( {1 - 40x + 400{x^2}} \right) - 1 + 20x + 10\\ & = 1200{x^2} - 100x + 12\end{align*}\]
    6. Compare the results of this to (e)! The order in which composition is written is important! Make sure you pay attention to the order.

      \[\begin{align*}\left( {g \circ f} \right)\left( x \right) & = g\left( {f\left( x \right)} \right)\\ & = g\left( {3{x^2} - x + 10} \right)\\ & = 1 - 20\left( {3{x^2} - x + 10} \right)\\ & = - 60{x^2} + 20x - 199\end{align*}\]