In Exercises graphexpfirsta - graphexplasta, sketch the graph of \(g\) by starting with the graph of \(f\) and using transformations. Track at least three points of your choice and the horizontal asymptote through the transformations. State the domain and range of \(g\).
\(f(x) = 2^{x}\), \(g(x) = 2^{x} - 1\)

Use the Desmos graph below with the settings \(f(x)=2^x,\,a=1,\,b=1,\,h=0,\,k=-1\)

domain: \((-\infty , \infty )\)

range: \((-1, \infty )\)

\(f(x) = \left (\frac {1}{3}\right )^{x}\), \(g(x) = \left (\frac {1}{3}\right )^{x-1}\)

Use the Desmos graph below with the settings \(f(x)=\left (\frac {1}{3}\right )^{x},\,a=1,\,b=1,\,h=1,\,k=0\)

domain: \((-\infty , \infty )\)

range: \((0, \infty )\)

\(f(x) = 3^{x}\), \(g(x) = 3^{-x}+2\)

Use the Desmos graph below with the settings \(f(x)=3^x,\,h=0,\,k=2,\,a=1,\,b=-1\)

domain: \((-\infty , \infty )\)

range: \((2, \infty )\)

\(f(x) = 10^{x}\), \(g(x) = 10^{\frac {x+1}{2}} - 20\)

Use the Desmos graph below with the settings \(f(x)=10^x,\,a=1,\,b=\frac {1}{2},\,h=-\frac {1}{2},\,k=-20\)

domain: \((-\infty , \infty )\)

range: \((-20, \infty )\)

\(f(t) = (0.5)^{t}\), \(g(t) = 100(0.5)^{0.1t}\)

Use the Desmos graph below with the settings \(f(t)=0.5^t,\,a=100,\,b=0.1,\,h=0,\,k=0\)

domain: \((-\infty , \infty )\)

range: \((0, \infty )\)

\(f(t) = (1.25)^{t}\), \(g(t) = 1 - (1.25)^{t-2}\)

Use the Desmos graph below with the settings \(f(t)=1.25^t,\,a=-1,\,b=1,\,h=2,\,k=1\)

domain: \((-\infty , \infty )\)

range: \((-\infty ,1)\)

\(f(t) = e^{t}\), \(g(t) = 8 - e^{-t}\)

Use the Desmos graph below with the settings \(f(t)=e^t,\,h=0,\,k=8,\,a=-1,\,b=-1\)

domain: \((-\infty , \infty )\)

range: \((-\infty ,8)\)

\(f(t) = e^{t}\), \(g(t) = 10e^{-0.1t}\)

Use the Desmos graph below with the settings \(f(t)=e^t,\,h=0,\,k=0,\,a=10,\,b=-0.1\)

domain: \((-\infty , \infty )\)

range: \((0,\infty )\)

In Exercises, expformfirsta - expformlasta, the graph of an exponential function is given. Find a formula for the function in the form \(F(x) = a \cdot 2^{bx-h}+k\).
Points: \(\left (-2, -\frac {5}{2} \right )\), \(\left (-1, -2 \right )\), \(\left (0, -1 \right )\),
Asymptote: \(y = -3\).

[Picture]

Use the Desmos graph below with the settings \(f(x)=2^x,\,a=1,\,b=1,\,h=-1,\,k=-3\)

Points: \(\left (-1, 1 \right )\), \(\left (0, 2 \right )\), \(\left (1, \frac {5}{2} \right )\), Asymptote: \(y = 3\).

Use the Desmos graph below with the settings \(f(x)=2^x,\,a=-1,\,b=-1,\,h=0,\,k=3\)

Points: \(\left (\frac {5}{2}, \frac {1}{2} \right )\), \(\left (3,1 \right )\), \(\left (\frac {7}{2}, 2 \right )\),
Asymptote: \(y = 0\).

[Picture]

Use the Desmos graph below with the settings \(f(x)=2^x,\,a=1,\,b=2,\,h=6,\,k=0\)

Points: \(\left (-\frac {1}{2}, 6 \right )\), \(\left (0,3 \right )\), \(\left (\frac {1}{2}, \frac {3}{2} \right )\),
Asymptote: \(y = 0\).

[Picture]

Use the Desmos graph below with the settings \(f(x)=2^x,\,a=3,\,b=-2,\,h=0,\,k=0\)

Find a formula for each graph in Exercises expformfirsta - expformlasta of the form \(G(x) = a \cdot 4^{bx-h} + k\). Did you change your solution methodology? What is the relationship between your answers for \(F(x)\) and \(G(x)\) for each graph?

You can use the same Desmos links in the solutions for Exercises expformfirsta - expformlasta to check your formulas for G(x).

Since \(2 = 4^{\frac {1}{2}}\), one way to obtain the formulas for \(G(x)\) is to use properties of exponents. For example, \(F(x) = 2^{x+1}-3 = \left (4^{\frac {1}{2}}\right )^{x+1} -3 = 4^{\frac {1}{2}(x+1)} - 3 = 4^{\frac {1}{2} x + \frac {1}{2}} - 3\). In order, the formulas for \(G(x)\) are:

  • \(G(x) = 4^{\frac {1}{2}x+\frac {1}{2}}-3\)
  • \(G(x) = -4^{-\frac {1}{2} x} + 3\)
  • \(G(x) = 4^{x-3}\)
  • \(G(x) =3 \cdot 4^{-x}\)
In Example expfcngraphsex number findformulaforexpexample, we obtained the solution \(F(x) = -2^{x+3} + 4\) as one formula for the given graph by making a simplifying assumption that \(a = -1\). This exercises explores if there are any other solutions for different choices of \(a\).
  1. Show \(G(x) = -4 \cdot 2^{x+1} + 4\) also fits the data for the given graph, and use properties of exponents to show \(G(x) = F(x)\). (Use the fact that \(4 = 2^2\) …)
  2. With help from your classmates, find solutions to Example expfcngraphsex number findformulaforexpexample using \(a = -8\), \(a = -16\) and \(a = -\frac {1}{2}\). Show all your solutions can be rewritten as: \(F(x) = -2^{x+3} + 4\).
  3. Using properties of exponents and the fact that the range of \(2^{x}\) is \((0, \infty )\), show that any function of the form \(f(x) = -a \cdot 2^{bx-h} + k\) for \(a> 0\) can be rewritten as \(f(x) = - 2^{c} \, 2^{bx-h} + k = -2^{bx-h+c} + k\). Relabeling, this means every function of the form \(f(x) = -a \cdot 2^{bx-h} + k\) with four parameters (\(a\), \(b\), \(h\), and \(k\)) can be rewritten as \(f(x) = - 2^{bx - H} + k\), a formula with just three parameters: \(b\), \(H\), and \(k\). Conclude that every solution to Example expfcngraphsex number findformulaforexpexample reduces to \(F(x) = -2^{x+3} + 4\) .
In Exercises decomposebasicexpfirst - decomposebasicexplast, write the given function as a nontrivial decomposition of functions as directed.
For \(f(x) = e^{-x} +1 \), find functions \(g\) and \(h\) so that \(f=g+h\).

One possible solution is \(g(x) = e^{-x}\) and \(h(x) = 1\).
For \(f(x) = e^{2x} - x\), find functions \(g\) and \(h\) so that \(f=g-h\).

One possible solution is \(g(x) = e^{2x}\) and \(h(x) = x\).
For \(f(t) = t^2 e^{-t}\), find functions \(g\) and \(h\) so that \(f=gh\).

One possible solution is \(g(t) = t^2\) and \(h(t) = e^{-t}\).
For \(r(x) = \dfrac {e^{x} - e^{-x}}{e^{x}+e^{-x}}\), find functions \(f\) and \(g\) so \(r = \dfrac {f}{g}\).

One possible solution is \(f(x) = e^{x} - e^{-x}\) and \(g(x) = e^{x}+e^{-x}\).
For \(k(x) = e^{-x^2}\), find functions \(f\) and \(g\) so that \(k = g \circ f\).

One possible solution is \(f(x) = -x^2\) and \(g(x) = e^{x}\).
For \(s(x) =\sqrt {e^{2x} - 1}\), find functions \(f\) and \(g\) so \(s = g \circ f\).

One possible solution is \(f(x) = e^{2x} -1\) and \(g(x) = \sqrt {x}\).

The amount of money in a savings account, \(A(t)\), in dollars, \(t\) years after an initial investment is made is given by: \(A(t) = 500(1.05)^{t}\), for \(t \geq 0\).

  1. Find and interpret \(A(0)\), \(A(1)\), and \(A(2)\).

    \(A(0) = 500\), so the principal is \(\$500\). \(A(1) = 525\), so after \(1\) year, there is \(\$525\) in the savings account. \(A(2) =551.25\), so after \(2\) years, there is \(\$551.25\) in the savings account.

  2. Find and interpret the relative rate of change of \(A\) over the intervals \([0,1]\), \([1,2]\), \([0,2]\).

    The relative rate of change of \(A\) over the intervals \([0,1]\) and \([1,2]\) is \(0.05\) which means the savings account is growing by \(5 \%\) each year for those two years. Over the interval \([0,2]\), the relative rate of change is \(0.1025\) meaning the account has grown by \(10.25 \%\) over the course of the first two years. Note this is greater than the sum of the two rates \(5 \% + 5 \% = 10 \%\). This is due to the ‘compounding effect’ and will be discussed in greater detail in Section ExpLogApplications.

  3. Find, simplify, and interpret the relative rate of change of \(A\) over the \([t, t+1]\). Assume \(t \geq 0\).

    The relative rate of change of \(A\) over the \([t, t+1]\) is \(0.05\). This means over the course of one year, the savings account grows by \(5 \%\).

  4. Use a graphing utility to estimate how long until the savings account is worth \(\$1500\). Round your answer to the nearest year.

    Graphing \(y= A(t)\) and \(y = 1500\), we find they intersect when \(t \approx 22.5\) so it takes approximately \(22-23\) years for the savings account to grow to \(\$1500\) in value.
Based on census data, the population of Lake County, Ohio, in 2010 was 230,041 and in 2015, the population was 229,437.

  1. Show the percentage change in the population from 2010 to 2015 is approximately \(-0.263 \%\).

    \(\frac {229437-230041}{230041} \approx 0.263 \%\).

  2. If this percentage change remains constant, predict the population of Lake County in 2020.

    Since 2020 is five years after 2015, we expect the population to decrease by \(0.263 \%\) of 229437, or approximately 603 people. Hence, we approximate the population in 2020 as 228834.

  3. Assuming this percentage change per five years remains constant, find an expression for the population \(P(t)\) of Lake County where \(t\) is the number of five year intervals after 2010. (So \(t = 0\) corresponds to 2010, \(t = 1\) corresponds to \(2015\), \(t = 2\) corresponds to \(2020\), etc.)

    Definitions expfcnpointbaseform and rrc and ensuing discussion on that page is useful here.

    \(P(t) = 230041(1-0.00263)^t = 230041(0.99737)^{t}\), \(t \geq 0\).

  4. Use your answer to populationfiveyear to predict the population of Lake County in the year 2017.

    Since 2017 is 7 years after 2010, we set \(t = \frac {7}{5} = 1.4\) and find \(P(1.4) \approx 229194\). So the population is approximately 229, 194 in 2017.

  5. Let \(A(t)\) represent the population of Lake County \(t\) years after 2010 where the we approximate the percentage change in population per year as \(-\frac {0.263 \%}{5} = -0.0526 \%\). Find a formula for \(A(t)\) and compare your predictions with \(A(t)\) to those given by \(P(t)\). In particular, what population does each model give for the year 2050? Discuss any discrepancies with your classmates.

    \(A(t) = 230041(1 - 0.0005626)^{t} = 230041 (0.999474)^{t}\), \(t \geq 0\). Since 2050 is 40 years after 2010, using the model \(P(t)\), we divide \(\frac {40}{5} = 8\) and find \(P(8) \approx 225,245\). On the other hand, \(A(40) \approx 225,250\). This is more than roundoff error. There is a compounding effect which makes the functions \(A(t)\) and \(P(t)\) different.
Show that the average rate of change of a function over the interval \([x, x+2]\) is average of the average rates of change of the function over the intervals \([x,x+1]\) and \([x+1, x+2]\). Can the same be said for the average rate of change of the function over \([x, x+3]\) and the average of the average rates of change over \([x, x+1]\), \([x+1, x+2]\), and \([x+2, x+3]\)? Generalize.
If \(f(x) = b^{x}\) where \(b>0\), \(b \neq 1\), show \(f(x_{0}+\Delta x) = f(x_{0}) b^{\Delta x}\).
Which is larger: \(e^{\pi }\) or \(\pi ^{e}\)? How do you know? Can you find a proof that doesn’t use technology?

  1. Use properties of exponential functions to show that if \(f(x) = e^{x}\), then

    \[ f'(x) = \lim _{h \rightarrow 0} \frac {f(x+h) - f(x)}{h} = \lim _{h \rightarrow 0} e^{x} \, \left ( \frac {e^{h} - 1}{h} \right )\]
  2. Numerically and graphically investigate the limit: \(\ds {\lim _{h \rightarrow 0}}\) \(\frac {e^{h} - 1}{h}\).
  3. Use parts diffquotex and specialexlimit to show to your surprise and delight that the derivative of \(e^{x}\) is …\(e^{x}\).

  4. Write the equations of the tangent lines to \(y = e^{x}\) at the following points:

    1. \((0,1)\)
    2. \((1, e)\)
    3. \(\left (-1, e^{-1} \right )\)