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MPE1 Practice Problems

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Math Placement Exam

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Directions. The following are review problems for the MPE1. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "More Examples" to see similar examples. While there are no videos for the MPE1 problems, the similar examples do have videos explaining the solution.
 
  1. Rationalize the denominator. \(\displaystyle \frac{14}{3+\sqrt{2}}\)

    \[
    \begin{align}
    \left(\dfrac{14}{3+\sqrt{2}}\right) \left(\dfrac{3-\sqrt{2}}{3-\sqrt{2}}\right)&=\dfrac{42-14\sqrt{2}}{9-2}\\
    &=\dfrac{7\left(6-2\sqrt{2}\right)}{7}\\
    &=6-2\sqrt{2}
    \end{align}
    \]
    Follow the link below to see more examples. 
    Section A.2 – Exponents and Radicals


  2. Find the sum or difference as indicated, and write your answer in simplified form.\[\frac{x+2a-3}{x+a}-\frac{x+6}{2x}\]

    \[\begin{align*}
    \frac{x+2a-3}{x+a}-\frac{x+6}{2x} &=\left(\frac{x+2a-3}{x+a}\right) \left(\frac{2x}{2x}\right) - \left(\frac{x+6}{2x}\right)\left(\frac{x+a}{x+a}\right)\\
    &=\frac{(x+2a-3)(2x)-(x+6)(x+a)}{(x+a)(2x)}\\
    &=\frac{2x^2+4xa-6x-x^2-6x-xa-6a}{(x+a)(2x)}\\
    &=\frac{x^2+3xa-12x-6a}{(x+a)(2x)}
    \end{align*}\]

    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions


  3. Factor and reduce to simplest form.\[\frac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2}\]

    \[ \dfrac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2} = \dfrac{(3x-2y)(2x+5y)}{(3x-2y)(x+4y)}=\dfrac{2x+5y}{x+4y}\]
    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions


  4. Simplify the following completely.\[\frac{\left(x^{-4}y^{2/5}\right)^{-3/4}}{x^{2/3}y^{-5/6}}\]

    \[\begin{align*}
    \frac{\left(x^{-4}y^{2/5}\right)^{-3/4}}{x^{2/3}y^{-5/6}}
    &=\frac{x^{12/4}y^{-6/20}}{x^{2/3}y^{-5/6}}\\
    &=x^{3-2/3}y^{-6/20-^-5/6}\\
    &=x^{7/3}y^{32/6}\\
    &=x^{7/3}y^{8/15}\\
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.2 – Exponents and Radicals


  5. Solve the following equation: \(5(x-7)-13(x-7)-6=0\).

    \[\begin{align*}
    5(x-7)-13(x-7)-6&=0\\
    (x-7)(5-13)-6&=0\\
    (x-7)(-8)-6&=0\\
    -8x+56-6&=0\\
    -8x+50&=0\\
    -8x&=-50\\
    x&=\dfrac{50}{8}
    \end{align*}\]


  6. Find the point \((x,y)\) which satisfies both equations. What is the value of \(x+y\)?\[\begin{align*} -2x+4y&=12\\ 3x-5y&=-3\end{align*}\]

    Multiply \(-2x+4y=12\) by \(3\) and \(3x-5y=-3\) by 2. This gives
    \[\begin{align*}
    -6x+12y&=36\\
    6x-10y&=-6\\
    \end{align*}\]
    Add these two equations and we get \(2y=30\) \(\enspace \Longrightarrow \enspace\) \(y=15\). Substituting \(y=15\) into \(3x-5y=-3\) gives \(x=24\). So the point that satisfies both equations is \((24,15)\), and the value of \(x+y\) is \(24+15 =39\). 
    Follow the link below to see more examples. 
    Section 7.1&2 – Systems of Equations


  7. Two investments are made, totaling \($10,000\). In one year, these investments yield \($650\) in simple interest. Part of the \($10,000\) is invested at \(5\frac{1}{2}\%\), and the rest at \(6\frac{3}{4}\%\). How much more money is invested at \(6\frac{3}{4}\%\)?

    Let \(x\) be the amount invested at \(5\frac{1}{2}\%\) and \(y\) be the amount invested at \(6\frac{3}{4}\%\). The resulting system of equations is 
    \[\begin{align*}
    x+y&=10,000\\
    0.055x+0.0675y&=650
    \end{align*}\]
    Solve the system of equations to find \(x\) and \(y\). 
    \(x=10,000-y\) \(\enspace\Longrightarrow\enspace\) \(0.055(10,000-y)+0.0675y=650\) \(\enspace\Longrightarrow\enspace\) \(550-0.055y+0.0675y=650\) \(\enspace\Longrightarrow\enspace\) \(0.0125y = 100\)  \(\enspace\Longrightarrow\enspace\) \(y=8000\). Since \(x+y=10,000\), we know that \(x=2000\), and thus \($6,000\) more is invested at \(6\frac{3}{4}\%\).
    Follow the link below to see more examples. 
    Section 7.1&2 – Systems of Equations


  8. Given the linear equation \(2ax + 3by = 7c\), where \(a\), \(b\) and \(c > 0\), if \(x\) decreases by \(10\) units, what is the corresponding change in \(y\)?

    Since \(2ax+3by=7c\) \(\enspace\Longrightarrow\enspace\) \(3by=7c-2ax\) \(\enspace\Longrightarrow\enspace\) \(y=\dfrac{7c-2ax}{3b}=\dfrac{-2a}{3b}x +\dfrac{7c}{3b}\). If \(x\) decreases by \(10\), then we know \(y=\dfrac{-2a}{3b}(x-10)+\dfrac{7c}{3b}=\dfrac{-2ax}{3b}+\dfrac{20a}{3b}+\dfrac{7c}{3b}\), so \(y\) will increase by \(\dfrac{20a}{3b}\).


  9.  Line \(A\) passes through the points \((2k+3, 4k−6)\) and \((−2, 16)\). Find the value of \(k\) if line \(A\) has a slope of \(0\).

    \(\dfrac{4k-6-16}{2k+3+2}=0\) \(\enspace\Longrightarrow\enspace\) \(\dfrac{4k-22}{2k+5}=0\) \(\enspace\Longrightarrow\enspace\) \(4k-22=0\) \(\enspace\Longrightarrow\enspace\) \(k=\dfrac{22}{4}=5.5\)
    Follow the link below to see more examples. 
    Section 1.3 – Linear Equations in Two Variables


  10. Jay wants to make a box, with no lid (or top), out of a \(10'' \times 6''\) rectangular piece of cardboard. If Jay cuts squares with dimensions \(x\) by \(x\) out of each corner of the cardboard, and then folds up the corners to make an open box, find a function that represents:
    1. The volume of the box.
    2. The surface area of the box.

      1. \(V=(10-2x)(6-2x)x=4x^3-32x^2+60x\)
      2.  \[\begin{align*} SA&=2x(6-2x)+2x(10-2x)+(6-2x)(10-2x)\\&=12x-4x^2+20x-4x^2+60-32x+4x^2\\&=-4x^2+60\end{align*}\]


  11. Solve for \(x\) in the inequality \(\dfrac{5x+2}{x-10}\geq 3\).

    \[\begin{align*}
    \dfrac{5x+2}{x-10}&\geq 3\\
    \dfrac{5x+2}{x-10}-3&\geq 0\\
    \dfrac{5x+2-3(x-10)}{x-10} &\geq 0\\
    \dfrac{5x+2-3x+30}{x-10}&\geq 0\\
    \dfrac{2x+32}{x-10}&\geq 0
    \end{align*}\]
    The expression \(\dfrac{2x+32}{x-10}\) will be greater than or equal to zero on \((-\infty, -16]\cup(10,\infty)\).
    Follow the link below to see more examples. 
    Section 2.7 – Nonlinear Inequalities


  12. Find the domain of the function below.\[f(x)=\dfrac{\sqrt{x^2-3x-4}}{6x^2-54}\]

    To find the domain of \( f(x)=\dfrac{\sqrt{x^2-3x-4}}{6x^2-54}\) we know \(x^2-3x-4\geq 0\) and \(6x^2-54 \neq 0\). \(x^2-3x-4\geq 0\) for \((-\infty, -1]\cup[4,\infty)\) and \(6x^2-54\neq 0\) for \((-\infty,-3)\cup(-3,3)\cup(3,\infty)\). So the domain of \(f(x)\) is \( (-\infty,-3)\cup (-3,-1]\cup [4,\infty).\)
    Follow the link below to see more examples. 
    Section 1.4 – Functions


  13. Find the domain of the function below. \[f(x)=\left\{\begin{array}{cc}\dfrac{2x^2+13}{x^2-1}, & x<0\\ \dfrac{5x-26}{x+2}, & x\geq 0\end{array} \right.\]

    For the function \(\dfrac{2x^2+13}{(x+1)(x-1)}\) on \( x<0\) the domain is \((-\infty,-1)\cup(-1,0)\). For the function \(\dfrac{5x-26}{x+2}\) on \(x\geq 0\) the domain is \([0,\infty)\). So for the function \(f(x)\), the domain is \( (-\infty,-1)\cup (-1,\infty)\).
    Follow the link below to see more examples. 
    Section 1.4 – Functions


  14. Find the \(x\)-intercept(s) of the function \(f(x)=\dfrac{6x^2-7x-5}{4x^2-12x-7}\), if any exist.

    \[\begin{align*}
    f(x)&=\dfrac{6x^2-7x-5}{4x^2-12x-7}\\
    &=\dfrac{(2x+1)(3x-5)}{(2x+1)(2x-7)}\\
    &=\dfrac{3x-5}{2x-7}
    \end{align*}\]
    The \(x\)-intercept is where \(3x-5=0\) \(\enspace\Longrightarrow\enspace\) \(x=\dfrac{5}{3}\).
    Follow the link below to see more examples. 
    Section 2.6 – Rational Functions


  15. Find the vertical and horizontal asymptote(s) of the function \( f(x)=\dfrac{6x^2-7x-5}{4x^2-12x-7}\), if any exist.

    \[\begin{align*}
    f(x)&=\frac{6x^2-7x-5}{4x^2-12x-6}\\
    &=\frac{(2x+1)(3x-5)}{(2x+1)(2x-7)}\\
    &=\frac{3x-5}{2x-7}
    \end{align*}\]
    The vertical asymptote is where \(2x-7=0\) so it is \(x=\dfrac{7}{2}\).

    The horizontal asymptote is at \(y= \dfrac{6}{4}\) which reduces to \(y= \dfrac{3}{2}\).
    Follow the link below to see more examples. 
    Section 2.6 – Rational Functions


  16. Find the \(x\)- and \(y\)-intercepts for the function \(f(x)=x^3-9x\).

    To find the \(x\)-intercepts solve \(f(x)=0\) \(\enspace \Longrightarrow\enspace \) \(x(x+3)(x-3)=0\) \(\enspace \Longrightarrow\enspace \) \(x=0,\pm 3\).

    To find the \(y\)-intercept, substitue \(x=0\) \(\enspace \Longrightarrow\enspace \) \(0^3-9(0)=0\). Hence, the \(y\)-intercept is \(y=0\).
     
    Follow the link below to see more examples. 
    Section 2.2 – Polynomial Functions of Higher Degree


  17. Find the domain of the following functions: 
    1. \(f(x)=\sqrt{-x^2-4x+5}\)
    2. \(g(t)=\ln(4t-3)\)
    3. \(h(x)=\dfrac{1}{x^3+3x^2-x-3}\)

      1. We must have \(−x^2−4x+5\geq 0\) for \(f(x)\) to be defined. Thus, \((−x+1)(x+5)\geq 0\) which occurs on \([−5, 1]\). Thus, the domain of \(f(x)\) is \([−5, 1]\).
      2. We must have \(4t − 3 > 0\) for \(g(t)\) to be defined. Thus, the domain for \(g(t)\) is \(\left(\dfrac{3}{4},\infty\right)\).
      3. We must have \(x^3+3x^2-x-3\neq 0\) for \(h(x)\) to be defined. \[\begin{align*} x^3+3x^2-x-3&\neq 0\\ x^2(x+3)-(x+3)&\neq 0\\  (x+3)\left(x^2-1\right) &\neq 0 \\(x+3)(x-1)(x+1)&\neq 0 \end{align*}\] So \(x\neq \pm 1\) and \(x\neq -3\). Thus, the domain of \(h(x)\) is \((-\infty, -3)\cup (-3,-1) \cup (-1,1) \cup (1,\infty)\).


  18. Simplify the expression \( \dfrac{2}{\sqrt{x^5}} \left(\sqrt[3]{4x}\right)\).

    \[\begin{align*}
    \left( \dfrac{2}{\sqrt{x^5}}\right) \left(\sqrt[3]{4x}\right) &= \left(2x^{-5/2} \right) (4x)^{1/3}\\
    &=\left(2x^{-5/2}\right) \left(4^{1/3}\cdot x^{1/3}\right) \\
    &=\left(2x^{-5/2}\right)\left( \left(2^2\right)^{1/3}x^{1/3}\right)\\
    &=\left(2x^{-5/2}\right)\left(2^{2/3}x^{1/3}\right)\\
    &=\dfrac{2^{5/3}}{x^{13/6}}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.2 – Exponents and Radicals


  19. If we begin with the graph of \(f(x) = x^2\) and shift \(f (x)\) 4 units to the right, shrink f (x) vertically by a factor of \(\frac{1}{2}\), and then shift \(f(x)\) upward 10 units, write the equation for the transformed graph.

    The transformed graph is \(g(x)=\dfrac{1}{2}(x-4)^2+10\).
    Follow the link below to see more examples. 
    Section 1.6&7 – Parent Functions and Transformations


  20. Solve the following equation for \(x\).\[\log(x+2)+\log(x-1)=1\]

    The domain of \(\log(x+2)\) is \((-2,\infty)\). The domain of \(\log(x-1)\) is \((1,\infty)\). So, the domain of \(\log(x+2)+\log(x-1)\) is \((1,\infty)\). Now, use properties of logarithms to solve the equation. 
    \[\begin{align*}
    \log(x+2)+\log(x-1)&=1\\
    \log\left[ (x+2)(x-1)\right] &= 1\\
    (x+2)(x-1)&=10^1\\
    x^2+x-2&=10\\
    x^2+x-12&=0\\
    (x+4)(x-3)&=0\\
    x&=-4,3
    \end{align*}\]
    Since \(x=-4\) is not in the domain, \(x=3\) is the only solution.
     
    Follow the link below to see more examples. 
    Section 3.4 – Exponential and Logarithmic Equations


  21. Factor the expression below completely. \[3x^2\left(4x^2+1\right)^8+64x^4\left(4x^2+1\right)^7\]

    \[\begin{align*}
    3x^2\left(4x^2+1\right)^8+64x^4\left(4x^2+1\right)^7 &= x^2\left(4x^2+1\right)^7 \left(3\left(4x^2+1\right)+64x^2\right)\\
    &=x^2\left(4x^2+1\right)^7\left(12x^2+3+64x^2\right)\\
    &=x^2\left(4x^2+1\right)^7\left(76x^2+3\right)
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.3 – Polynomials and Factoring


  22. How far from the base of an \(18\) foot tall pole must a person be standing if the angle of elevation from the ground to the pole is \(41^\circ\)?

    Refer to the figure below. We know that \(\tan (41^\circ)=\dfrac{18}{x}\). So, \(x=\dfrac{18}{\tan 41^\circ}\) or \(x=18\cot(41^\circ)\). So, the person must be \(18\cot(41^\circ)\) feet from the base of the pole.

    Follow the link below to see more examples. 
    Section 4.3 – Right Triangle Trigonometry


  23. Find \(f\circ g\) (also denoted \(f(g(x))\)) if \(f(x)=\dfrac{x}{x+1}\) and \(g(x)=\dfrac{2}{x}\). Simplify.

    \[\begin{align*}
    f\circ g &= \frac{ \frac{2}{x}}{\frac{2}{x}+1}\\
    &= \frac{ \frac{2}{x}}{\frac{2}{x}+\frac{x}{x}}\\
    &=\frac{ \frac{2}{x}}{\frac{2+x}{x}}\\
    &=\frac{2}{x}\cdot \frac{x}{2+x} \\
    &=\frac{2}{2+x}
    \end{align*}\]


  24. Perform the indicated operations and simplify. \[\frac{8}{x+1}-\left(\frac{y}{z+2} \div \frac{y-4}{w}\right)\]

    \[\begin{align*}
    \frac{8}{x+1}-\left( \frac{y}{z+2} \div \frac{y-4}{w}\right) &= \frac{8}{x+1}-\left( \frac{y}{z+2} \cdot \frac{w}{y-4}\right)\\
    &=\frac{8}{x+1} - \left( \frac{yw}{(z+2)(y-4)}\right) \\
    &=\frac{ 8(z+2)(y-4)-yw(x+1)}{(x+1)(z+2)(y-4)}\\
    &=\frac{8zy-32z+16y-64-ywx-yw}{(x+1)(z+2)(y-4)}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions


  25. Solve the equation \(e^{2x}-2e^x-3=0\) for \(x\).

    First, factor the equation: \(e^{2x}-2e^x-3=(e^x)^2-2e^x-3=(e^x-3)(e^x+1)\). Solving \((e^x-3)(e^x+1)=0\) \(\enspace \Longrightarrow\enspace \) \(e^x-3=0\) or \(e^x+1=0\) \(\enspace \Longrightarrow\enspace \) \(e^x=3\) or \(e^x=-1\). But \(e^x\) will never be negative, so the only solution is \(e^x=3\) \(\enspace \Longrightarrow\enspace \) \(x=\ln(3)\). 
    Follow the link below to see more examples. 
    Section 3.4 – Exponential and Logarithmic Equations


  26. Find the equation of the line passing through the point \((5,1)\) with a slope of 7. Use the equation you find to determine the value of \(y\) when \(x=-4.\)

    Using the point-slope equation, we get \(y-1=7(x-5)\) \(\enspace \Longrightarrow\enspace \) \(y=7x-35+1\) \(\enspace \Longrightarrow\enspace \) \(y=7x-34\). Using this to find \(y\) when \(x=-4\) gives \(y=7(-4)-34\) \(\enspace \Longrightarrow\enspace \) \(y=-62\). 
    Follow the link below to see more examples. 
    Section 1.3 – Linear Equations in Two Variables


  27. If \(f(x)=\sqrt{x+4}\), find and simplify \(\dfrac{f(2+h)-f(2)}{h}\).

    \[\begin{align*}
    \dfrac{f(2+h)-f(2)}{h} &=\dfrac{\sqrt{2+h+4}-\sqrt{2+4}}{h}\\
    &=\left(\dfrac{\sqrt{6+h}-\sqrt{6}}{h}\right) \cdot \left( \dfrac{\sqrt{6+h}+\sqrt{6}}{\sqrt{6+h}+\sqrt{6}}\right)\\
    &=\dfrac{6+h-6}{h\left(\sqrt{6+h}+\sqrt{6}\right)}\\
    &=\dfrac{1}{\sqrt{6+h}+\sqrt{6}}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section 1.4 – Difference Quotient


  28. Simplify \(\dfrac{ \left(x^2y^4\right)^5\left( x^3y\right)^{-3}}{xy}\).

    \[\begin{align*}
    \dfrac{ \left(x^2y^4\right)^5\left( x^3y\right)^{-3}}{xy}
    &=\dfrac{x^{10}y^{20}x^{-9}y^{-3}}{xy}\\
    &=\dfrac{xy^{17}}{xy}\\
    &=y^{16}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.2 – Exponents and Radicals


  29. Simplify \(\sqrt[3]{\left(a^3b\right)}\sqrt[3]{64a^4b^2}\).

    \(\sqrt[3]{\left(a^3b\right)}\sqrt[3]{64a^4b^2}=\sqrt[3]{64a^7b^3}=4a^2b\left(\sqrt[3]{a}\right)\)
    Follow the link below to see more examples. 
    Section A.2 – Exponents and Radicals


  30. Perform the operations indicated and simplify.\[\frac{x^2}{x^2-x-2}-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}\]

    \[\begin{align*}
    \frac{x^2}{x^2-x-2}&-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}\\
    &=\frac{x^2}{(x-2)(x+1)}-\frac{4}{(x+3)(x-2)}+\frac{x}{(x+1)(x+3)}\\
    &=\left(\frac{x+3}{x+3}\right)\left(\frac{x^2}{(x-2)(x+1)}\right) -\left(\frac{x+1}{x+1}\right) \left(\frac{4}{(x+3)(x-2)}\right)+\left(\frac{x-2}{x-2}\right)\left(\frac{x}{(x+1)(x+3)}\right)\\
    &= \frac{x^3+3x^2-4x-4+x^2-2x}{(x+3)(x-2)(x+1)}\\
    &=\frac{x^3+4x^2-6x-4}{(x+3)(x-2)(x+1)}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions


  31. Find all zeros, horizontal, and vertical asymptotes for \(f(x)=\dfrac{3x^2-14x-5}{4x^2-17x-15}\).

    Factor and simplify: \(f(x)=\dfrac{3x^2-14x-5}{4x^2-17x-15}=\dfrac{(3x+1)(x-5)}{(4x+3)(x-5)}=\dfrac{3x+1}{4x+3}\). This yields a zero of \(x=-\dfrac{1}{3}\), a vertical asymptote of \(x=-\dfrac{3}{4}\), and a horizontal asymptote of \(y=\dfrac{3}{4}\).
    Follow the link below to see more examples. 
    Section 2.6 – Rational Functions


  32. If \(\theta\) is in Quadrant II, and \(\sin(\theta) = \dfrac{1}{7}\), what is \(\cos(\theta)\)?

    Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), we find that \(\cos^2 \theta = 1-\sin^2\theta=1-\dfrac{1}{49}=\dfrac{48}{49}\). This gives us \(\cos \theta= \pm \sqrt{\dfrac{48}{49}}=\pm \dfrac{\sqrt{48}}{7} = \pm \dfrac{4\sqrt{3}}{7}\). Since \(\theta\) is in quadrant II, \(\cos \theta <0\), hence \(\cos \theta = -\dfrac{4\sqrt{3}}{7}\). 
    Follow the link below to see more examples. 
    Section 4.4 – Trigonometric Functions of Any Angle


  33. Use the properties of logarithms to expand the expression \( \ln \left( \dfrac{\sqrt{x}y^5}{(z+1)^4}\right)\).

    We know \(\ln(ab)=\ln a + \ln b\), \( \ln \left(\dfrac{a}{b}\right)=\ln a - \ln b\), and \( \ln a^b =b \ln a\). Using these properties, we obtain
    \[\begin{align*}
    \ln \left( \frac{ \sqrt{x}y^5}{(z+1)^4} \right) &= \ln \left( \sqrt{x}y^5\right) - \ln (z+1)^4\\
    &=\ln \left(x^{1/2}y^5\right)-4\ln(z+1)\\
    &=\ln x^{1/2}+\ln y^5 -4\ln(z+1)\\
    &=\dfrac{1}{2}\ln x+5\ln y -4\ln(z+1)
    \end{align*}\]
    Follow the link below to see more examples. 
    Section 3.3 – Properties of Logarithms


  34. Evaluate \( \sec \dfrac{2\pi}{3} - \tan \dfrac{\pi}{6}\).

    \(\sec \dfrac{2\pi}{3}-\tan \dfrac{\pi}{6} = -2 - \dfrac{1}{\sqrt{3}}\)
    Follow the link below to see more examples.
    Section 4.2 – Trigonometric Functions: The Unit Circle 


  35. If we begin with a rectangle with length 5 inches and width 4 inches, then increase the length by \(8 \%\), what is the change in area?

    We begin with a rectangle of area \(20\) in\(^2\). If we increase the length by \(8 \%\), then our new length is \(5.4\) inches. This gives a new area of \(21.6\) in\(^2\). This gives a total change in area of \(1.6\) in\(^2\). 


  36. Evaluate \( f(2)-f(-3)\) given \( f(x)=\left\{ \begin{array}{ll} x^3+1, & x>1\\2x^2-3,&x\leq 1\end{array}\right.\)

    \(f(2)=(2)^3+1=9\) and \(f(-3)=2(-3)^2-3=18-3=15\), so \(f(2)-f(-3)=9-15=-6\). 
    Follow the link below to see more examples.
    Section 1.4 – Functions


  37. Simplify the expression \(\dfrac{\cos^2 (\theta)}{1+\sin(\theta)}\).

    Use the identity \(\cos^2\theta = 1-\sin^2 \theta\) to get 
    \[\begin{align*}
    \frac{\cos^2(\theta)}{1+\sin(\theta)} &= \frac{1-\sin^2\theta}{1+\sin \theta}\\
    &=\frac{(1+\sin \theta)(1-\sin \theta)}{1+\sin\theta}\\
    &=1-\sin \theta
    \end{align*}\]
    Follow the link below to see more examples.
    Section 5.2 – Verifying Trigonometric Identities


  38. Evaluate \(\log_4\left(\dfrac{1}{\sqrt[3]{16}}\right)\).

    \[\begin{align*}
    \log_4 \left(\dfrac{1}{\sqrt[3]{16}}\right) &=\log_4\left( \frac{1}{\sqrt[3]{4^2}}\right)\\
    &=\log_4 \left( \dfrac{1}{4^{2/3}}\right)\\
    &=\log_4 4^{-2/3} \\
    &=-\dfrac{2}{3}\\
    \end{align*}\]

    Follow the link below to see more examples.
    Section 3.2 – Logarithmic Functions and Their Graphs


  39. Simplify \(\dfrac{\frac{1}{a}-b}{\frac{1}{b^3}+a}\).

    \(\dfrac{\frac{1}{a}-b}{\frac{1}{b^3}+a} =
    \dfrac{\left(\frac{1}{a}-b\right)\left(ab^3\right)}{\left(\frac{1}{b^3}+a\right)\left(ab^3\right)}
    =\dfrac{b^3-ab^4}{a+a^2b^3}\)
    Follow the link below to see more examples.
    Section A.4 – Rational Expressions


  40. A bacteria contains \(1200\) bacteria and doubles every day. How many hours will it take the culture to reach 10000 bacteria?

    Use the exponential model \(y(t)=y_0e^{kt}\), where \(y(t)\) is the size of the population at time \(t\) and \(y_0\) is the initial size of the population. We know that \(y_0=1200\), hence \(y(t)=1200e^{kt}\).​ Since the population doubles everyday, we know
    \[\begin{align*}
    2400&=1200e^{k\cdot 24}\\
    2&=e^{24k}\\
    \ln 2&= 24 k\\
    k&=\dfrac{1}{24}\ln 2\\
    k&= \ln 2^{1/24}\\
    \end{align*}\]
    Thus, \(y(t)=1200e^{t\ln 2^{1/24}} = 1200 e^{\ln 2^{t/24}}=1200\cdot 2^{t/24}\) (note \(t\) is in hours). Now solve for \(t\) when \(y(t)=10000\)
    \[\begin{align*}
    10000&=1200\cdot 2^{t/24}\\
    \dfrac{10000}{1200}&=2^{t/24}\\
    \dfrac{25}{3}&=2^{t/24}\\
    \log_2\left(\dfrac{25}{3}\right) &= \dfrac{t}{24}\\
    t&=24\log_2 \left(\dfrac{25}{3}\right)\\
    \end{align*}\]
    So, it will take \(24 \log_2 \left( \dfrac{25}{3}\right)\) hours for the culture to reach \(10,000\) bacteria.
    Follow the link below to see more examples.
    Section 3.5 – Exponential and Logarithmic Models