In Exercises powergraphexfirst - powergraphexlast, sketch the graph of the new function by starting with the graph of the given function and using Theorem linearrationalpowergraphs. Track at least two points and state the domain and range using interval notation.
\(F(x) = (x-2)^{\frac {2}{3}}-1\)

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

Domain: \((-\infty , \infty )\), Range: \([-1, \infty )\)

\(G(t) = (t+3)^{\pi } +1\)

Use the Desmos graph below with the settings \(g(t)=t^\pi ,\,h=-3,\,k=1,\,a=1,\,b=1\)

Domain: \([-3, \infty )\), Range: \([1, \infty )\)

\(F(x) = 3-x^{\frac {2}{3}}\)

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

Domain: \((-\infty , \infty )\), Range: \((-\infty ,3]\)

\(G(t) = (1-t)^{\pi }-2\)

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

Domain: \((-\infty ,1]\), Range: \([-2, \infty )\)

\(F(x) =(2x+5)^{\frac {2}{3}}+1\)

Use the Desmos graph below with the settings \(f(x)=x^\frac {2}{3},\,h=-5,\,k=1,\,a=-1,\,b=2\)

Domain: \((-\infty , \infty )\), Range: \([1, \infty )\)

\(G(t) = \left ( \frac {t+3}{2}\right )^{\pi }-1\)

Use the Desmos graph below with the settings \(g(t)=t^\pi ,\,h=-\frac {3}{2},\,k=-1,\,a=1,\,b=\frac {1}{2}\)

Domain: \([-3, \infty )\), Range: \([-1, \infty )\)

In Exercises findformulaforpowergraphfirst - findformulaforpowergraphlast, find a formula for each function below in the form \(F(x) = a(bx-h)^{\frac {2}{3}}+k\).

NOTE: There may be more than one solution!

\(y=F(x)\)

[Picture]

One solution is: \(F(x) = 2(x-1)^{\frac {2}{3}}-2\)
\(y = F(x)\) [Picture]

One solution is: \(F(x) =-(x+1)^{\frac {2}{3}} + 4\)
For each function in Exercises powerfcngraphexfirst - powerfcngraphexlast below

  1. Analytically:

    • find the domain.
    • find the axis intercepts.
    • analyze the end behavior.

  2. Graph the function with help from a graphing utility and determine:

    • the range.
    • the local extrema, if they exist.
    • intervals of increase/decrease.
    • any ‘unusual steepness’ or ‘local’ verticality.
    • vertical asymptotes.
    • horizontal / slant asymptotes.
  3. Construct a sign diagram for each function using the intercepts and graph.
  4. Comment on any observed symmetry.
\(f(x) = x^{\frac {2}{3}}(x - 7)^{\frac {1}{3}}\)

\(f(x) = x^{\frac {2}{3}}(x - 7)^{\frac {1}{3}}\)
Domain: \((-\infty , \infty )\)
Intercepts: \((0,0)\), \((7,0)\)
Graph:
\[\graph {f(x)=x^{2/3}*(x-7)^{1/3}}\]

As \(x \rightarrow -\infty \), \(f(x) \rightarrow -\infty \)
As \(x \rightarrow \infty \), \(f(x) \rightarrow \infty \)
Range: \((-\infty , \infty )\)
Local minimum: \(\approx (4.667, -3.704)\)
Local maximum: \((0,0)\) (this is a cusp)
Increasing: \((-\infty , 0]\), \(\approx [4.667, \infty )\)
Decreasing: \([0, 4.667]\)
Unusual steepness at \(x = 7\)
Using Calculus it can be shown that \(y = x - \frac {7}{3}\) is a slant asymptote of this graph.
Sign Diagram:

\(f(x) = x^{\frac {3}{2}}(x - 7)^{\frac {1}{3}}\)

\(f(x) = x^{\frac {3}{2}}(x - 7)^{\frac {1}{3}}\)
Domain: \([0, \infty )\)
Intercepts: \((0,0)\), \((7,0)\)
Graph:
\[\graph {f(x)=x^{3/2}*(x-7)^{1/3}}\]

As \(x \rightarrow \infty \), \(f(x) \rightarrow \infty \)
Range: \(\approx [-14.854, \infty )\)
Local minimum: \(\approx (5.727, -14.854)\)
Increasing: \(\approx [5.727, \infty )\)
Decreasing: \(\approx [0, 5.727]\)
Unusual steepness at \(x = 7\)
Sign Diagram:

\(g(t) = 2t(t+3)^{-\frac {1}{3}}\)

\(g(t) = 2t(t+3)^{-\frac {1}{3}}\)
Graph:
\[\graph {g(t)=2t(t+3)^{-1/3}}\]

Domain: \((-\infty , -3) \cup (-3, \infty )\)
Intercept: \((0,0)\)
As \(t \rightarrow \pm \infty \), \(g(t) \rightarrow \infty \)
Range: \((-\infty , \infty )\)
Local minimum: \(\approx (-4.5, 7.862)\)
Increasing: \(\approx [-4.5, -3)\), \((-3,\infty )\)
Decreasing: \(\approx (-\infty , -4.5]\)
Vertical Asymptote: \(t = -3\)
Sign Diagram:

\(g(t) = t^{\frac {3}{2}}(t-2)^{-\frac {1}{2}}\)

\(g(t)= t^{\frac {3}{2}}(t-2)^{-\frac {1}{2}}\)
Domain: \((2, \infty )\)
As \(t \rightarrow \infty \), \(g(t) \rightarrow \infty \)
Graph:
\[\graph {g(t)=t^{3/2}(t-2)^{-\frac {1}{2}}}\]

Range: \(\approx [5.196, \infty )\)
Local minimum: \(\approx (3, 5.196)\)
Increasing: \(\approx [3, \infty )\)
Decreasing: \(\approx (2,3]\)
Vertical asymptote: \(t = 2\)
Using Calculus it can be shown that \(y = t+1\) is a slant asymptote of this graph.
Sign Diagram:

\(f(x) = x^{0.4} (3-x)^{0.6}\)

\(f(x) = x^{0.4}(3-x)^{0.6}\)
Domain: \((-\infty , \infty )\)
Intercepts: \((0,0)\), \((3,0)\)
Graph:
\[\graph {f(x) = x^{0.4}(3-x)^{0.6}}\]

As \(x \rightarrow -\infty \), \(f(x) \rightarrow \infty \)
As \(x \rightarrow \infty \), \(f(x) \rightarrow -\infty \)
Range: \((-\infty , \infty )\)
Local minimum: \((0,0)\) (this is a cusp)
Local maximum: \(\approx (1.2, 1.531)\)
Increasing: \(\approx [0, 1.2]\)
Decreasing: \(\approx (-\infty , 0]\), \([1.2, \infty )\)
Unusual Steepness: \(x = 3\)
Sign Diagram:

\(f(x) = x^{0.5} (3-x)^{0.5}\)

\(f(x) = x^{0.5}(3-x)^{0.5}\)
Graph:
\[\graph {f(x) = x^{0.5}(3-x)^{0.5}}\]

Domain: \([0,3]\)
Intercepts: \((0,0)\), \((3,0)\)
Range: \(\approx [0, 1.5]\)
Increasing: \(\approx [0, 1.5]\)
Decreasing: \(\approx [1.5, 3]\)
Unusual Steepness: \(x=0\), \(x = 3\)

\(g(t) = 4t (9-t^2)^{-\sqrt {2}}\)

\(g(t) = 4t (9-t^2)^{-\sqrt {2}}\)
Graph:
\[\graph {g(t)=4t(9-t^2)^{-\sqrt {2}}}\]

Domain: \((-3, 3)\)
Intercepts: \((0,0)\)
Range: \((-\infty , \infty )\)
Increasing: \((-3,3)\)
Sign Diagram:

\(g(t) = 3(t^2+1)^{-\pi }\)

\(g(t) = 3(t^2+1)^{-\pi }\)
Domain: \((-\infty , \infty )\)
Graph:
\[\graph {g(t)=3(t^2+1)^{-\pi }}\]

Intercept: \((0,3)\)
As \(t \rightarrow \pm \infty \), \(g(t) \rightarrow 0\)
Range: \((0, 3]\)
Increasing: \((-\infty , 0]\)
Decreasing: \([0, \infty )\)
Horizontal asymptote: \(y =0\)
Sign Diagram:

For each function \(f(x)\) listed below, compute the average rate of change over the indicated interval. What trends do you observe? How do your answers manifest themselves graphically? Compare the results of this exercise with those of Exercise monomialarcexercise in Section GraphsofPolynomials and Exercise laurentarcexercise in Section IntroRational
\[ \begin{array}{|r||c|c|c|c|} \hline f(x) & [0.9, 1.1] & [0.99, 1.01] &[0.999, 1.001] & [0.9999, 1.0001] \\ \hline x^{\frac {1}{2}} &&&& \\ \hline x^{\frac {2}{3}} &&&& \\ \hline x^{-0.23} &&&& \\ \hline x^{\pi } &&&& \\ \hline \end{array} \]

As in Exercise monomialarcexercise in Section GraphsofPolynomials and Exercise laurentarcexercise in Section IntroRational, the slopes of these curves near \(x = 1\) approach the value of the exponent on \(x\).
\[ \begin{array}{|r||c|c|c|c|} \hline f(x) & [0.9, 1.1] & [0.99, 1.01] &[0.999, 1.001] & [0.9999, 1.0001] \\ \hline x^{\frac {1}{2}} & 0.5006 & \approx \frac {1}{2} & \approx \frac {1}{2} & \approx \frac {1}{2} \\ \hline x^{\frac {2}{3}} & 0.6672 & 0.6667 & \approx \frac {2}{3} & \approx \frac {2}{3} \\ \hline x^{-0.23} & -0.2310 & \approx -0.23 &\approx -0.23 & \approx -0.23 \\ \hline x^{\pi } &3.1544 & 3.1417 & \approx \pi & \approx \pi \\ \hline \end{array} \]
The National Weather Service uses the following formula to calculate the wind chill:
\[ W = 35.74 + 0.6215 \, T_{a} - 35.75\, V^{0.16} + 0.4275 \, T_{a} \, V^{0.16} \]
where \(W\) is the wind chill temperature in \(^{\circ }\)F, \(T_{a}\) is the air temperature in \(^{\circ }\)F, and \(V\) is the wind speed in miles per hour. Note that \(W\) is defined only for air temperatures at or lower than \(50^{\circ }\)F and wind speeds above \(3\) miles per hour.

  • Suppose the air temperature is \(42^{\circ }\) and the wind speed is \(7\) miles per hour. Find the wind chill temperature. Round your answer to two decimal places.

    \(W \approx 37.55^{\circ }\)F.

  • Suppose the air temperature is \(37^{\circ }\)F and the wind chill temperature is \(30^{\circ }\)F. Find the wind speed. Round your answer to two decimal places.

    \(V \approx 9.84\) miles per hour.
As a follow-up to Exercise WindChillTemperature, suppose the air temperature is \(28^{\circ }\)F.

  • Use the formula from Exercise WindChillTemperature to find an expression for the wind chill temperature as a function of the wind speed, \(W(V)\).

    \(W(V) = 53.142 - 23.78 V^{0.16}\). Since we are told in Exercise WindChillTemperature that wind chill is only effect for wind speeds of more than 3 miles per hour, we restrict the domain to \(V > 3\).

  • Solve \(W(V) = 0\), round your answer to two decimal places, and interpret.

  • Graph the function \(W\) using a graphing utility and check your answer to part WindChill0.

Suppose Fritzy the Fox, positioned at a point \((x,y)\) in the first quadrant, spots Chewbacca the Bunny at \((0,0)\). Chewbacca begins to run along a fence (the positive \(y\)-axis) towards his warren. Fritzy, of course, takes chase and constantly adjusts his direction so that he is always running directly at Chewbacca. If Chewbacca’s speed is \(v_1\) and Fritzy’s speed is \(v_2\), the path Fritzy will take to intercept Chewbacca, provided \(v_2\) is directly proportional to, but not equal to, \(v_1\) is modeled by
\[ y = \frac {1}{2} \left (\frac {x^{1+ v_{1}/v_{2}}}{1+v_1/v_2}- \frac {x^{1-v_1/v_2}}{1-v_1/v_2}\right ) + \frac {v_1 v_2}{v_2^2-v_1^2} \]

  • Determine the path that Fritzy will take if he runs exactly twice as fast as Chewbacca; that is, \(v_2 = 2v_1\). Use your calculator to graph this path for \(x \geq 0\). What is the significance of the \(y\)-intercept of the graph?

  • Determine the path Fritzy will take if Chewbacca runs exactly twice as fast as he does; that is, \(v_1 = 2v_2\). Use a graphing utility to graph this path for \(x > 0\). Describe the behavior of \(y\) as \(x \rightarrow 0^{+}\) and interpret this physically.

    • With the help of your classmates, generalize parts (a) and (b) to two cases: \(v_2 > v_1\) and \(v_2 < v_1\). We will discuss the case of \(v_1 = v_2\) in Exercise pursuitlog in Section ExpLogApplications.