Solution.

1. One way to proceed to solve \((7-x)^{\frac {3}{2}} = 8\) is to use Definition rationalexponentdefna to rewrite \((7-x)^{\frac {3}{2}}\) as either \((\sqrt {7-x})^3\) or \(\sqrt {(7-x)^3}\). We opt for the former since, thinking ahead, \(8\) is a perfect cube:

   \[ \begin{array}{rclr} (7-x)^{\frac {3}{2}} & = & 8 & \\ (\sqrt {7-x})^3 & = & 8 & \text {use } a^{\frac {m}{n}} = \left (\sqrt [n]{a}\right )^m = \sqrt [n]{a^m}\\ \sqrt [3]{(\sqrt {7-x})^3} & = & \sqrt [3]{8} & \text {extract cube roots} \\ \sqrt {7-x} & = & 2 & \text {$\sqrt [3]{u^3}= u$} \\ \end{array} \]

   From \(\sqrt {7-x} = 2\), we square both sides and obtain \(7-x = 4\), so \(x = 3\). We verify our answer analytically by substituting \(x=3\) into the original equation and it checks.

   Geometrically, we are looking for where the graph of \(f(x) = (7-x)^{\frac {3}{2}}\) intersects the graph of \(g(x) = 8\). While we could sketch both curves by hand and gauge the reasonableness of the result, ^[1] (consider this an exercise!) we are instructed to use a graphing utility. Using desmos, we see the graphs intersect at one point, \((3,8)\), thereby checking our solution \(x = 3\).

2. Proceeding similarly to the above, to solve \((2t-1)^{\frac {2}{3}} -4 = 0\), we rewrite \((2t-1)^{\frac {2}{3}}\) as \((\sqrt [3]{2t-1})^2\) and solve:

   \[ \begin{array}{rclr} (2t-1)^{\frac {2}{3}} -4 & = & 0 & \\ (\sqrt [3]{2t-1})^2 - 4 & = & 0 & \text {rewrite using the defintion of rational exponents} \\ (\sqrt [3]{2t-1})^2 & = & 4 & \text {isolate the variable term} \\ \sqrt {(\sqrt [3]{2t-1})^2 } & = & \sqrt {4} & \text {extract square roots} \\ |\sqrt [3]{2t-1}| & = & 2 & \text {$\sqrt {u^2} = |u|$} \\ \sqrt [3]{2t-1} & = & \pm 2 & \text { for $c>0$, $|u| = c$ is equivalent to $u = \pm c$.} \end{array} \]

   From \(\sqrt [3]{2t-1} = 2\) we cube both sides and obtain \(2t-1 = 8\), so \(t = \frac {9}{2} = 4.5\). Similarly, from \(\sqrt [3]{2t-1} = -2\), we cube both sides and obtain \(2t-1 = -8\), so \(t = -\frac {7}{2} = -3.5\). Both of these solutions check in the given equation.

   In this case we are looking for where the graph of \(f(t) = (2t-1)^{\frac {2}{3}} -4\) intersects the graph of \(g(t) = 0\) - i.e., the \(t\)-intercepts of the graph of \(g\). We find these are \((-3.5,0)\) and \((4.5,0)\), as verified below using desmos.

3. Since \(0.5 = \frac {1}{2}\), we may rewrite \((x+3)^{0.5} = 2(7-x)^{0.5}+1\) as \((x+3)^{\frac {1}{2}} = 2(7-x)^{\frac {1}{2}}+1\). Using Definition rationalexponentdefna, we then have \(\sqrt {x+3} = 2\sqrt {7-x} + 1\). Since one of the square roots is already isolated, we can rid ourselves of it by squaring both sides.

   \[ \begin{array}{rclr} \sqrt {x+3} & = & 2\sqrt {7-x} + 1 & \\ (\sqrt {x+3})^2 & = & (2\sqrt {7-x} + 1)^2 & \text {square both sides} \\ x+3 & = & (2 \sqrt {7-x})^2 + 2 (2 \sqrt {7-x})(1) + 1 & \text {$(\sqrt {u})^2 = u$ and $(a+b)^2 = a^2 + 2ab +b^2$} \\ x+3 & = & 4(7-x) + 4\sqrt {7-x} + 1 & \text {$(ab)^2 = a^2b^2$ and, again, $(\sqrt {u})^2 = u$} \\ x+3 & = & 28-4x+4\sqrt {7-x} + 1 & \\ 5x-26 & = & 4\sqrt {7-x} & \text {isolate $\sqrt {7-x}$} \\ \end{array} \]

   We square both sides again and get \((5x-26)^2 = (4\sqrt {7-x})^2\) which reduces to \(25x^2-260x+676 = 16(7-x)\). At last, we have a quadratic equation which we can solve by setting to zero and factoring. We get \(25x^2-244x+564 = 0\), so \((x-6)(25x-94) = 0\) so \(x = 6\) or \(x = \frac {94}{25} = 3.76\). When we go to check these answers, we find \(x=6\) does check, but \(x = 3.76\) does not. Hence, \(x=3.76\) is an ‘extraneous’ solution. ^[2] (We invite the reader to see at which point in our machinations \(x=3.76\) does check. Knowing a solution is extraneous is one thing; understanding how it came about is another.)

   We graph both \(f(x) = \sqrt {x+3}\) and \(g(x) = 2\sqrt {7-x} + 1\) below using desmos and confirm there is only one intersection point, \((6,3)\).

4. While we could approach solving \(2t^{\frac {2}{3}} + 5t^{\frac {1}{3}} = 3\) as the previous example, we would encounter cubing binomials ^[3] (that is, expanding things like \((a+b)^3\).) which we would prefer to avoid. Instead, we take a step back and notice there are three terms here with the exponent on one term, \(t^{\frac {2}{3}}\) exactly twice the exponent on another term, \(t^{\frac {1}{3}}\). We have ourselves a ‘quadratic in disguise.’ ^[4] (See Section AppQuadEqus or, more recently, Example polyquadinform in Section RealZeros.) To help us see the forest for the trees, we let \(u = t^{\frac {1}{3}}\) so that \(u^2 = t^{\frac {2}{3}}\). (Note that since root here, \(3\), is odd, we can use the properties of exponents stated in Theorem exponentprops.) Hence, in terms of \(u\), the equation \(2t^{\frac {2}{3}} + 5t^{\frac {1}{3}} = 3\) becomes the quadratic \(2u^2 + 5u = 3\), or \(2u^2 + 5u - 3 = 0\). Factoring gives \((2u-1)(u+3) = 0\) so \(u = t^{\frac {1}{3}} = \frac {1}{2}\) or \(u = t^{\frac {1}{3}} = -3\). Since \(t^{\frac {1}{3}} = \sqrt [3]{t}\), we solve both equations by cubing both sides to get \(t = \frac {1}{8} = 0.125\) and \(t = -27\). Both of these solutions check in our original equation.

   Using desmos, we graph \(f(t) = 2t^{\frac {2}{3}} + 5t^{\frac {1}{3}}\) and \(g(t) = 3\). We see the two graphs intersect at \((0.125,3)\), and, after some adjustment, we find the other intersection point at \((-27,3)\).

5. Next we are to solve \(2(3x-1)^{-0.5} = 3x (3x-1)^{-1.5}\) which, when written without negative exponents is: \(\frac {2}{(3x-1)^{0.5}} = \frac {3x}{(3x-1)^{1.5}}\). Since the rational exponents here are \(0.5 = \frac {1}{2}\) and \(1.5 = \frac {3}{2}\), both involve an even indexed root (the square root in this case!) which means \(3x-1 \geq 0\). Moreover, since the \(3x-1\) resides in the denominator \(3x - 1 \neq 0\) so our equation is really valid only for values of \(x\) where \(3x-1>0\) or \(x > \frac {1}{3}\). Hence, we clear denominators and can apply Theorem exponentprops:

   \[ \begin{array}{rclr} \frac {2}{(3x-1)^{0.5}} & = & \frac {3x}{(3x-1)^{1.5}} & \\ \left [ \frac {2}{(3x-1)^{0.5}} \right ] \cdot (3x-1)^{1.5} & = & \left [ \frac {3x}{(3x-1)^{1.5}} \right ] \cdot (3x-1)^{1.5} & \\ 2 \cdot \frac {(3x-1)^{1.5}}{(3x-1)^{0.5}} & = & 3x & \\ 2 (3x-1)^{1.5-0.5} & = & 3x & \text {we can subtract exponents since $3x-1 > 0$.} \\ 2 (3x-1)^{1} & = & 3x & \\ \end{array} \]

   We get \(6x-2 = 3x\), or \(x = \frac {2}{3}\). Since \(x = \frac {2}{3} > \frac {1}{3}\), we keep it and, sure enough, it checks in our original equation. Graphically we see \(f(x)=2(3x-1)^{-0.5}\) intersects \(g(x) = 3x (3x-1)^{-1.5}\) at the point \((0.6667, 2)\) which is the desmos’s way of representing \(\left (\frac {2}{3}, 2\right )\).

6. Our last equation to solve is \(6(9-t^2)^{\frac {1}{3}} = 4t^2 (9-t^2)^{-\frac {2}{3}}\), which, when rewritten without negative exponents is: \(6(9-t^2)^{\frac {1}{3}} = \frac {4t^2}{(9-t^2)^{\frac {2}{3}}}\) . Again, the root here (\(3\)) is odd, so we can use the exponent properties listed in Theorem exponentprops. We begin by clearing denominators:

   \[ \begin{array}{rclr} 6(9-t^2)^{\frac {1}{3}} & = & \frac {4t^2}{(9-t^2)^{\frac {2}{3}}} & \\ 6(9-t^2)^{\frac {1}{3}} \cdot (9-t^2)^{\frac {2}{3}} & = & \left [ \frac {4t^2}{(9-t^2)^{\frac {2}{3}}} \right ] \cdot (9-t^2)^{\frac {2}{3}} & \\ 6(9-t^2)^{\frac {1}{3} + \frac {2}{3}}& = & 4t^2 & \text {we can add exponents since the root here $3$ is odd.} \\ 6(9-t^2)^{1} & = & 4t^2 &\\ \end{array} \]

   We get \(54 - 6t^2 = 4t^2\) or \(10t^2 = 54\). As fraction \(t^2 = \frac {54}{10} = \frac {27}{5}\) so \(t = \pm \sqrt {\frac {27}{5}} = \pm 3 \sqrt {15}{5}\). While not the easiest to check analytically, both of these solutions do work in the original equation. Graphing \(f(t) = 6(9-t^2)^{\frac {1}{3}} \) and \(g(t) = 4t^2 (9-t^2)^{-\frac {2}{3}}\) below using desmos, we see the graphs intersect when \(t \approx \pm 2.324\) which are decimal approximations of our exact answers.

1. For 1910, the recorded labor and capital figures for the US Economy are \(144 \%\) and \(208 \%\) of the 1899 figures, respectively. Find \(P\) using these figures and interpret your answer.
2. The recorded production value figure for 1920 is \(231 \%\) of the 1899 figure. Use this to write \(K\) as a function of \(L\), \(K = f(L)\). Find and interpret \(f(193)\).
3. Graph \(K = f(L)\) and interpret the behavior as \(L \rightarrow 0^{+}\) and \(L \rightarrow \infty \).

Solution.

1. In this case, \(P = 1.01 L^{0.75} K^{0.25} = 1.01 (144)^{0.75} (208)^{0.25} \approx 159\) which means the dollar value of the total US Production in 1920 was approximately \(159 \%\) of what it was in 1899. ^[6] (This answer is remarkably accurate. Note: all the dollar values here are recorded in ‘1880’ dollars, per the source article. )

2. We are given \(P = 231 = 1.01L^{0.75} K^{0.25}\), so to write \(K\) as a function of \(L\), we need to solve this equation for \(K\). Since \(L\) and \(K\) are positive by definition, we can employ properties of exponents:

   \[ \begin{array}{rclr} 231 & = & 1.01 L^{0.75} K^{0.25} & \\ \frac {231}{1.01 L^{0.75}} & = & \frac {1.01 L^{0.75} K^{0.25}}{1.01 L^{0.75}} & \text {$L>0$, hence $L^{0.75} \neq 0$.}\\ K^{0.25} & = & 228.\overline {7128}L^{-0.75} & \text {rewrite} \\ \left (K^{0.25}\right )^{\frac {1}{0.25}} & = & \left (228.\overline {7128}L^{-0.75}\right )^{\frac {1}{0.25}} & \\ K^{\frac {0.25}{0.25}} & = & (228.\overline {7128})^{\frac {1}{0.25}} L^{-\frac {0.75}{0.25}} & \text {for a power of a power, multiply exponents} \\ K & = & (228.\overline {7128})^{4} L^{-3} & \text {simplify} \\ \end{array} \]

   Hence, \(K = f(L) = (228.\overline {7128})^{4} L^{-3}\) where \(L>0\). We find \(f(193) = (228.\overline {7128})^{4} (193)^{-3} \approx 381\) meaning that in order to maintain a production level of \(231 \%\) of 1889 with a labor level at \(193 \%\) of 1889, the required capital is \(381 \%\) that of 1889. ^[7] (The actual recorded figure is 407.)

3. The function \(f(L)\) is a Laurent Monomial (see Section IntroRational) with \(n = 3\) and \(a = (228.\overline {7128})^{4}\) and is graphed below on the applied domain, \(L>0\).

   

   We see that as \(L \rightarrow 0^{+}\), \(f(L) \rightarrow \infty \). This means that in order to maintain the given production level, as the available labor diminishes, the capital requirement becomes unbounded. As \(L \rightarrow \infty \), we have \(f(L) \rightarrow 0\) meaning that as the available labor increases, the need for capital diminishes. The graph of \(f\) is called an ‘isoquant’ - meaning ‘same quantity.’ In this context, the graph displays all combinations of labor and capital, \((L,K)\) which result in the same production level, in this case, \(231 \%\) of what was produced in 1889.

□

Solve the following inequalities. Check your answers graphically with a calculator.

1. \(2-\sqrt [4]{x+3} \geq 0\)
2. \(t^{2/3} < t^{4/3} - 6\)
3. \(3 (2-x)^{\frac {1}{3}} \leq x (2-x)^{-\frac {2}{3}}\)
4. \((t-4)^{\frac {2}{3}} \geq -\frac {2t}{3(t-4)^{\frac {1}{3}}}\)

Solution.

1. To solve \(2-\sqrt [4]{x+3} \geq 0\), it is tempting to rewrite this inequality as \(2 \geq \sqrt [4]{x+3}\) and rid ourselves of the fourth root by raising both sides of this inequality to the fourth power. While this technique works sometimes, it doesn’t work all the time since raising both sides of an inequality to the fourth (more generally, to an even) power does not necessarily preserve inequalities . ^[9] (For instance, \(-2 \leq 1\) but \((-2)^{4} \geq (1)^2\). We invite the reader to see what goes wrong if attempting to solve either of the following inequalities using this method: \(-2 \geq \sqrt [4]{x+3}\), which has no solution, or \(-2 \leq \sqrt [4]{x+3}\), whose solution is \([-3,\infty )\).) For that reason, we solve this inequality using a sign diagram since this technique will always produce a correct solution.

   We already have all the nonzero terms on one side of the inequality, so we let \(r(x) = 2-\sqrt [4]{x+3}\) and proceed to make a sign diagram. Owing to the presence of the fourth root, we know \(x+3 \geq 0\) or \(x \geq -3\). Hence, we only concern ourselves with the portion of the number line representing \([3, \infty )\). Next, we find the zeros of \(r\) by solving \(r(x) = 2-\sqrt [4]{x+3}=0\). We get \(\sqrt [4]{x+3} = 2\), so \(x+3= 16\) and we get \(x=13\). We find this solution checks in our original equation, ^[10] (Recall that raising both sides to an even power could produce extraneous solutions, so it is important we check here.) and proceed to construct the sign diagram below.

   

   Since we are looking for where \(r(x) = 2-\sqrt [4]{x+3} \geq 0\), we are looking for the zeros of \(r\) along with the intervals over which \(r(x)\) is \((+)\). We record our answer as \([-3, 13]\). Using desmos, we graph \(y = 2-\sqrt [4]{x+3}\), and we can see that, indeed, the graph is above the \(x\)-axis (\(y=0\)) from \([-3, 13)\) and meets the \(x\)-axis at \(x=13\), verifying our answer.

2. To solve \(t^{\frac {2}{3}} < t^{\frac {4}{3}} - 6\), we first rewrite as \(t^{\frac {4}{3}} -t^{\frac {2}{3}} - 6 > 0\). We set \(r(t) = t^{\frac {4}{3}} -t^{\frac {2}{3}} - 6\) and note that since the denominators in the exponents are \(3\), they correspond to cube roots, which means the domain of \(r\) is \((-\infty , \infty )\). To find the zeros for the sign diagram, we set \(r(t) = 0\) and attempt to solve \(t^{\frac {4}{3}} - t^{\frac {2}{3}} - 6 = 0\). Since there are three terms, and the exponent on one of the variable terms, \(t^{\frac {4}{3}}\), is exactly twice that of the other, \(t^{\frac {2}{3}}\), we have ourselves a ‘quadratic in disguise.’ If we let \(u = t^{\frac {2}{3}}\), then \(u^2 = t^{\frac {4}{3}}\), so in terms of \(u\), we have \(u^2 - u - 6 = 0\). Solving we get \(u = -2\) or \(u = 3\), hence \(t^{\frac {2}{3}} = -2\) or \(t^{\frac {2}{3}} = 3\). In root-power notation, these are \(\sqrt [3]{t^2} = -2\) or \(\sqrt [3]{t^2}= 3\). Cubing both sides of these equations results in \(t^2 = -8\), which admits no real solution, or \(t^2 = 27\), which gives \(t = \pm 3 \sqrt {3}\). Using these zeros, we construct the sign diagram below.

   

   We find \(r(t) = t^{\frac {4}{3}} -t^{\frac {2}{3}} - 6 > 0\) on \(\left (-\infty , -3 \sqrt {3}\right )\cup \left (3 \sqrt {3}, \infty \right )\). To check our answer graphically, we set \(f(t) = t^{\frac {2}{3}}\) and \(g(t) = t^{\frac {4}{3}}-6\). The solution to \(t^{\frac {2}{3}} < t^{\frac {4}{3}} - 6\) corresponds to the inequality \(f(t) < g(t)\), which means we are looking for the \(t\) values for which the graph of \(f\) is below the graph of \(g\). Using desmos, we see the graph of \(y = f(t) = t^{\frac {2}{3}}\) is below the graph of \(y = g(t) = t^{\frac {4}{3}}-6\) for \(t < - 5.196\) and again for \(t > 5.196\), which are desmos’s approximations to \(\pm 3 \sqrt {3}\).

3. To solve \(3 (2-x)^{\frac {1}{3}} \leq x (2-x)^{-\frac {2}{3}}\), we first gather all the nonzero terms to one side and obtain \(3 (2-x)^{\frac {1}{3}} - x (2-x)^{-\frac {2}{3}} \leq 0\). Setting \(r(x) = 3 (2-x)^{\frac {1}{3}} - x (2-x)^{-\frac {2}{3}}\), we note since the denominators of the rational exponents are odd, we have no domain concerns owing to even indexed roots. However, the negative exponent on the second term indicates a denominator.

   Rewriting \(r(x)\) with positive exponents, we obtain

   \[r(x) = 3 (2-x)^{\frac {1}{3}} - \frac {x}{(2-x)^{\frac {2}{3}}}\]
   Setting the denominator equal to zero we get \((2-x)^{\frac {2}{3}} = 0\), which reduces to \(2-x=0\), or \(x=2\). Hence, the domain of \(r\) is \((-\infty , 2) \cup (2, \infty )\).

   To find the zeros of \(r\), we set \(r(x) = 0\), so we set about solving

   \[3 (2-x)^{\frac {1}{3}} - \frac {x}{(2-x)^{\frac {2}{3}}} = 0.\]
   Clearing denominators, we get \(3 (2-x)^{\frac {1}{3}} (2-x)^{\frac {2}{3}} - x = 0\). Since the denominators of the exponents are odd, we may use Theorem exponentprops to simplify this to \(3(2-x)^{1} - x = 0\), and obtain \(6-4x = 0\) or \(x = \frac {3}{2}\). In order for us to be able to more easily determine the sign of \(r(x)\) at the test values, we rewrite \(r(x)\) as a single term. ^[11] (This also gives us a chance to review some good intermediate algebra!) There are two schools of thought on how to proceed, so we demonstrate both.

   • Factoring Approach. From \(r(x) = 3 (2-x)^{\frac {1}{3}} - x (2-x)^{-\frac {2}{3}}\), we note that the quantity \((2-x)\) is common to both terms. When we factor out common factors, we factor out the quantity with the smaller exponent. In this case, since \(-\frac {2}{3} < \frac {1}{3}\), we factor \((2-x)^{-\frac {2}{3}}\) from both quantities. While it may seem odd to do so, we need to factor \((2-x)^{-\frac {2}{3}}\) from \((2-x)^{\frac {1}{3}}\), which results in subtracting the exponent \(-\frac {2}{3}\) from \(\frac {1}{3}\). We proceed using the usual properties of exponents.

     \[ \begin{array}{rclr} r(x) & = & 3 (2-x)^{\frac {1}{3}} - x (2-x)^{-\frac {2}{3}} & \\ & = & (2-x)^{-\frac {2}{3}} \left [ 3 (2-x)^{\frac {1}{3} - \left (-\frac {2}{3}\right )} - x\right ] & \\ & = & (2-x)^{-\frac {2}{3}}\left [3(2-x)^{\frac {3}{3}} - x\right ] & \\ & = & (2-x)^{-\frac {2}{3}}\left [3(2-x)^{1} - x\right ] & \\ & = & (2-x)^{-\frac {2}{3}}\left (6-4x\right ) & \\ & = & (2-x)^{-\frac {2}{3}}\left (6-4x\right ) & \\ \end{array}\]

     Written without negative exponents, we have \(r(x) = \frac {6-4x}{(2-x)^{\frac {2}{3}}}\).

   • Common Denominator Approach. We rewrite

     \[ \begin{array}{rclr} r(x) & = & 3 (2-x)^{\frac {1}{3}} - x (2-x)^{-\frac {2}{3}} & \\ & = & 3 (2-x)^{\frac {1}{3}} - \frac {x}{(2-x)^{\frac {2}{3}}} & \\ & = & \frac {3 (2-x)^{\frac {1}{3}}(2-x)^{\frac {2}{3}}}{(2-x)^{\frac {2}{3}}} - \frac {x}{(2-x)^{\frac {2}{3}}} & \text {common denominator} \\ & = & \frac {3 (2-x)^{\frac {1}{3} + \frac {2}{3}}}{(2-x)^{\frac {2}{3}}} - \frac {x}{(2-x)^{\frac {2}{3}}} & \text {root here, 3, is odd, so add exponents} \\ & = & \frac {3 (2-x)^{\frac {3}{3}}}{(2-x)^{\frac {2}{3}}} - \frac {x}{(2-x)^{\frac {2}{3}}} & \\ & = & \frac {3 (2-x)^1}{(2-x)^{\frac {2}{3}}} - \frac {x}{(2-x)^{\frac {2}{3}}} & \\ & = & \frac {3 (2-x) - x}{(2-x)^{\frac {2}{3}}} & \\ & = & \frac {6-4x}{(2-x)^{\frac {2}{3}}} & \\ \end{array}\]

   Using either approach, we end up with the same, simpler, expression for \(r(x)\) and we use that to create our sign diagram as shown below.

   

   We find \(r(x) \leq 0\) on \(\left [\frac {3}{2},2\right ) \cup (2, \infty )\). To check, we graph \(f(x)=3 (2-x)^{\frac {1}{3}}\) and \(g(x) = x (2-x)^{-\frac {2}{3}}\) using desmos. We confirm that the graphs intersect at \(x=\frac {3}{2}\) and the graph of \(y = f(x) = 3 (2-x)^{\frac {1}{3}}\) is below the graph of \(y = g(x) = x (2-x)^{-\frac {2}{3}}\) for \(x > \frac {3}{2}\), with the exception of \(x=2\) where it appears the graph of \(g\) has a vertical asymptote.

4. While it may be tempting to begin solving our last inequality by clearing denominators, owing to the odd root, the quantity \(3(t-4)^{\frac {1}{3}}\) can be both positive and negative for different values of \(t\). This means that if we chose to multiply both sides of our inequality by this quantity, we have no guarantee if the inequality would be preserved. Hence we proceed as usual by gathering all the nonzero terms to one side, and, with the ultimate goal of creating a sign diagram, get common denominators.

   \[ \begin{array}{rclr} (t-4)^{\frac {2}{3}} & \geq & - \frac {2t}{3(t-4)^{\frac {1}{3}}} & \\ (t-4)^{\frac {2}{3}} + \frac {2t}{3(t-4)^{\frac {1}{3}}} & \geq & 0 \\ \frac { (t-4)^{\frac {2}{3}} \cdot 3(t-4)^{\frac {1}{3}} }{3(t-4)^{\frac {1}{3}}} + \frac {2t}{3(t-4)^{\frac {1}{3}}} & \geq & 0 & \text {common denominator} \\ \frac {3(t-4)^{\frac {2}{3}+\frac {1}{3}}}{3(t-4)^{\frac {1}{3}}} + \frac {2t}{3(t-4)^{\frac {1}{3}}} & \geq & 0 & \text {root is odd so add exponents} \\ \frac {3(t-4)^{1}}{3(t-4)^{\frac {1}{3}}} + \frac {2t}{3(t-4)^{\frac {1}{3}}} & \geq & 0 & \\ \frac {3(t-4) + 2t}{3(t-4)^{\frac {1}{3}}} & \geq & 0 & \\ \frac {5t-12}{3(t-4)^{\frac {1}{3}}} & \geq & 0 & \\ \end{array} \]

   We identify \(r(t)\) as the left hand side of the inequality and see right away we must exclude \(t=4\) from the domain owing to the quantity \((t-4)\) in the denominator. As we have already mentioned, the root here (\(3\)) is odd, so we have no domain issues stemming from that. To find the zeros of \(r\), we set \(r(t) = 0\) which quickly reduces to solving \(5t-12 = 0\). We get \(t = \frac {12}{5}\). Our sign diagram is below.

   Show Alt Text

   Sign diagram showing the intervals on which the function \(r(x)\) is positive or negative. Dashed vertical line at \(x=2\) indicates a vertical asymptote. The function is positive on \((-\infty ,\frac {12}{5})\), negative on \((\frac {12}{5},4)\), and positive on \((4,\infty )\). The function has a zero at \(x=\frac {12}{5}\).
   

   From the sign diagram, we find \(r(t) \geq 0\) on \(\left (-\infty , \frac {12}{5} \right ] \cup (4, \infty )\). Graphing \(f(t) = (t-4)^{\frac {2}{3}}\) and \(g(t) = -\frac {2t}{3(t-4)^{\frac {1}{3}}}\) using desmos, we see the graph of \(y = f(t) = (t-4)^{\frac {2}{3}}\) is above the graph of \(y =g(t) = -\frac {2t}{3(t-4)^{\frac {1}{3}}}\) for \(t < 2.4\) and again for \(t > 4\), with an intersection point at \(t=2.4 = \frac {12}{5}\).

□