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    Home / Algebra Trig Review / Exponentials & Logarithms / Basic Exponential Functions
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    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    Paul
    August 27, 2020

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    Basic Exponential Functions

    First, let’s recall that for \(b > 0\) and \(b \ne 1\) an exponential function is any function that is in the form

    \[f\left( x \right) = {b^x}\]

    We require \(b \ne 1\) to avoid the following situation,

    \[f\left( x \right) = {1^x} = 1\]

    So, if we allowed \(b = 1\) we would just get the constant function, 1.

    We require \(b > 0\) to avoid the following situation,

    \[f\left( x \right) = {\left( { - 4} \right)^x}\hspace{0.25in}\hspace{0.25in} \Rightarrow \,\hspace{0.25in}\,\,\,\,f\left( {\frac{1}{2}} \right) = {\left( { - 4} \right)^{\frac{1}{2}}} = \sqrt { - 4} \]

    By requiring \(b > 0\) we don’t have to worry about the possibility of square roots of negative numbers.

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    1. Evaluate \(f\left( x \right) = {4^x}\), \(g\left( x \right) = {\left( {\displaystyle \frac{1}{4}} \right)^x}\) and \(h\left( x \right) = {4^{ - x}}\) at \(x = - 2, - 1,0,1,2\).
      Show Solution

      The point here is mostly to make sure you can evaluate these kinds of functions. So, here’s a quick table with the answers.

        \(\displaystyle x = - 2\) \(\displaystyle x = - 1\) \(\displaystyle x = 0\) \(\displaystyle x = 1\) \(\displaystyle x = 2\)
      \(\displaystyle f\left( x \right)\) \(\displaystyle f\left( { - 2} \right) = \frac{1}{{16}}\) \(\displaystyle f\left( { - 1} \right) = \frac{1}{4}\) \(\displaystyle f\left( 0 \right) = 1\) \(\displaystyle f\left( 1 \right) = 4\) \(\displaystyle f\left( 2 \right) = 16\)
      \(\displaystyle g\left( x \right)\) \(\displaystyle g\left( { - 2} \right) = 16\) \(\displaystyle g\left( { - 1} \right) = 4\) \(\displaystyle g\left( 0 \right) = 1\) \(\displaystyle g\left( 1 \right) = \frac{1}{4}\) \(\displaystyle g\left( 2 \right) = \frac{1}{{16}}\)
      \(\displaystyle h\left( x \right)\) \(\displaystyle h\left( { - 2} \right) = 16\) \(\displaystyle h\left( { - 1} \right) = 4\) \(\displaystyle h\left( 0 \right) = 1\) \(\displaystyle h\left( 1 \right) = \frac{1}{4}\) \(\displaystyle h\left( 2 \right) = \frac{1}{{16}}\)

      Notice that the last two rows give exactly the same answer. If you think about it that should make sense because,

      \[g\left( x \right) = {\left( {\frac{1}{4}} \right)^x} = \frac{{{1^x}}}{{{4^x}}} = \frac{1}{{{4^x}}} = {4^{ - x}} = h\left( x \right)\]
    2. Sketch the graph of \(f\left( x \right) = {4^x}\), \(g\left( x \right) = {\left( {\displaystyle \frac{1}{4}} \right)^x}\) and \(h\left( x \right) = {4^{ - x}}\) on the same axis system.
      Show Solution

      Note that we only really need to graph \(f\left( x \right)\) and \(g\left( x \right)\) since we showed in the previous Problem that \(g\left( x \right) = h\left( x \right)\). Note as well that there really isn’t too much to do here. We found a set of values in Problem 1 so all we need to do is plot the points and then sketch the graph. Here is the sketch,

    3. List as some basic properties for \(f\left( x \right) = {b^x}\).
      Show Solution

      Most of these properties can be seen in the sketch in the previous Problem.

      (a) \(f\left( x \right) = {b^x} > 0\) for every \(x\). This is a direct consequence of the requirement that \(b > 0\).

      (b) For any \(b\) we have \(f\left( 0 \right) = {b^0} = 1\).

      (c) If \(b > 1\) (\(f\left( x \right) = {4^x}\) above, for example) we see that \(f\left( x \right) = {b^x}\) is an increasing function and that,

      \[f\left( x \right) \to \infty {\mbox{ as }}x \to \infty \hspace{0.5in}{\rm{and}}\hspace{0.5in}f\left( x \right) \to 0{\mbox{ as }}x \to - \infty \]

      (d) If \(0 < b < 1\) (\(g\left( x \right) = {\left( {\frac{1}{4}} \right)^x}\) above, for example) we see that \(f\left( x \right) = {b^x}\) is an decreasing function and that,

      \[f\left( x \right) \to 0{\mbox{ as }}x \to \infty \hspace{0.5in}{\rm{and}}\hspace{0.5in}f\left( x \right) \to \infty {\mbox{ as }}x \to - \infty \]

      Note that the last two properties are very important properties in many Calculus topics and so you should always remember them!

    4. Evaluate \(f\left( x \right) = {{\bf{e}}^x}\), \(g\left( x \right) = {{\bf{e}}^{ - x}}\) and \(h\left( x \right) = 5{{\bf{e}}^{1 - 3x}}\) at \(x = - 2, - 1,0,1,2\).
      Show Solution

      Again, the point of this problem is to make sure you can evaluate these kinds of functions. Recall that in these problems \(e\) is not a variable it is a number! In fact, \({\bf{e}} = 2.718281828 \ldots \)

      When computing \(h\left( x \right)\) make sure that you do the exponentiation BEFORE multiplying by 5.

        \(\displaystyle x = - 2\) \(\displaystyle x = - 1\) \(\displaystyle x = 0\) \(\displaystyle x = 1\) \(\displaystyle x = 2\)
      \(\displaystyle f\left( x \right)\) \(\displaystyle 0.135335\) \(\displaystyle 0.367879\) \(\displaystyle 1\) \(\displaystyle 2.718282\) \(\displaystyle 7.389056\)
      \(\displaystyle g\left( x \right)\) \(\displaystyle 7.389056\) \(\displaystyle 2.718282\) \(\displaystyle 1\) \(\displaystyle 0.367879\) \(\displaystyle 0.135335\)
      \(\displaystyle h\left( x \right)\) \(\displaystyle 5483.166\) \(\displaystyle 272.9908\) \(\displaystyle 13.59141\) \(\displaystyle 0.676676\) \(\displaystyle 0.033690\)

    5. Sketch the graph of \(f\left( x \right) = {{\bf{e}}^x}\) and \(g\left( x \right) = {{\bf{e}}^{ - x}}\).
      Show Solution

      As with the other “sketching” problem there isn’t much to do here other than use the numbers we found in the previous example to make the sketch. Here it is,

      Note that from these graphs we can see the following important properties about \(f\left( x \right) = {{\bf{e}}^x}\) and \(g\left( x \right) = {{\bf{e}}^{ - x}}\).

      \[\begin{align*} & {{\bf{e}}^x} \to \infty {\mbox{ as }}x \to \infty & \hspace{0.5in} & {\rm{and}} & \hspace{0.5in} & {{\bf{e}}^x} \to 0{\mbox{ as }}x \to - \infty \\ & {{\bf{e}}^{ - x}} \to 0{\mbox{ as }}x \to \infty & \hspace{0.25in} & {\rm{and}} & \hspace{0.25in} & {{\bf{e}}^{ - x}} \to \infty {\mbox{ as }}x \to - \infty \end{align*}\]

      These properties show up with some regularity in a Calculus course and so should be remembered.

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