Math

Chem

Chem 7.31 - 7.45

Exam 2 - Annette Storms

1. Given $f(x) =\frac{5x+1}{x^2-9}$ and $g(x)=\frac{4x-2}{x^2-9}$

(a) Find $f+g$ and the domain.

$f+g =\frac{5x+1}{x^2-9} + \frac{4x-2}{x^2-9}$

$\color{blue}{f+g=\frac{9x-1}{x^2-9}}$

$x^2-9\ne0$
$x^2\ne9$
$x\ne\pm\sqrt{9}$
$x\ne\pm3$

The domain of $f+g$ is: $\color{blue}{\left(-\infty \:,\:-3\right)\cup \left(-3,\:3\right)\cup \left(3,\:\infty \:\right)}$

(b) Find $f \cdot g$ and the domain, leave in factored form.

$f\cdot g =\frac{5x+1}{x^2-9} \times \frac{4x-2}{x^2-9}$

$\color{blue}{f\cdot g =\frac{(5x+1)(4x-2)}{(x^2-9)(x^2-9)}}$

$(x^2-9)(x^2-9)=0$

$(x^2-9)^2=0$

Take the square root of both sides:

$x^2-9=0$

$x^2=9$

Take the square root of both sides:

$x=\pm\sqrt{9}$

$x=\pm3$

The domain of $f\cdot g$ is: $\color{blue}{\left(-\infty \:,\:-3\right)\cup \left(-3,\:3\right)\cup \left(3,\:\infty \:\right)}$

(c) Find $f/g$ and the domain.

$f/g = \frac{5x+1}{x^2-9} \div \frac{4x-2}{x^2-9}$

$f/g = (5x+1) \div (4x-2)$

$\color{blue}{f/g = \frac{5x+1}{4x-2}}$

$4x-2\ne0$

$4x\ne2$

$x\ne2/4$

$x\ne\frac{1}{2}$

The domain of $f/g$ is: $\color{blue}{\left(-\infty \:,\:\frac{1}{2}\right)\cup \left(\frac{1}{2},\:\infty \:\right)}$

2. Find $(f\circ g)(x)$ and the domain of $f\circ g$

$f(x)=\frac{2}{x+3}\:\:\:$ and $\:\:\:g(x)=\frac{1}{x}$

$(f \circ g)(x) = f(g(x))$

$f(g(x)) = f(\frac{1}{x})$

$f(\frac{1}{x}) = \frac{2}{\frac{1}{x}+3}$

$= 2 \div (\frac{1}{x}+3)$

$= 2 \div \frac{1+3x}{x}$

$= \frac{2}{1} \times \frac{x}{1+3x}$

$\color{blue}{f \circ g = \frac{2x}{1+3x}}$

$1+3x\ne0\\3x\ne-1\\x\ne-\frac{1}{3}$

The domain of $f \circ g$ is: $\color{blue}{\left(-\infty \:,\:-\frac{1}{3}\right)\cup \left(-\frac{1}{3},\:\infty \:\right)}$

3. For the quadratic function, $f(x)=-2x^2-12x+3$ ,

(a) Determine, without graphing, whether the function has a minimum or a maximum value. Explain.

The function $f(x)\\$ has a maximum value, because the leading coefficient is negative.

(b) Find the minimum or maximum value and determine where it occurs. Give answer as an ordered pair.

Apply the vertex formula, $-b/2a$ to find $x$ :

$x=-(-12)/(2(-2))\\=12/-4\\=-3$

Plug $-3$ in for $x$ to find $y$ :

$y=-2(-3)^2-12(-3)+3\\=-18+36+3\\=21$

Vertex occurs at the maximum, $\color{blue}{(-3,21)}$ .

The domain is the set of all real numbers:

Domain: $\color{blue}{(-\infty,\infty)}$

Since the vertex is a maximum and occurs at $y=21$ , the range is limited to all real numbers $\leq21$ :

Range: $\color{blue}{(-\infty,21]}$

4. Given the polynomial function,
$f(x)=x^3+7x^2-4x-28$ ,

(a) find the zeros and give the multiplicity of each zero.

$x^3+7x^2-4x-28=0$

$x^3+7x^2\:\:=\:\:x^2(x+7)$

$-4x-28\:\:=\:\:-4(x+7)$

$x^2(x+7) - 4(x+7)$

Factor out the common term $(x+7)$ :

$(x+7)(x^2 - 4)$

Factor $(x^2 - 4)$ :

$(x+2)(x-2)$

$(x+7)(x+2)(x-2)$

The zeros of the function are:
$\color{blue}{x=-7}$ (multiplicty of $\color{blue}{1}$ )
$\color{blue}{x=-2}$ (multiplicty of $\color{blue}{1}$ )
$\color{blue}{x=2}$ (multiplicty of $\color{blue}{1}$ )

(b) State whether the graph crosses the $x$ -axis, or touches it and turns around, at each zero.

Since the multiplicities of each zero is odd, the graph crosses the $x$ -axis at each zero.

5. Evaluate the following expressions

(a) $\frac{2x^5-8x^4+2x^3+x^2}{2x^3+1}$

        x² -4x  +1     
       ____________________
2x³+1 | 2x⁵−8x⁴ +2x³ +x² +0x +0
      -(2x⁵          +x²)
        -----------------
           −8x⁴ +2x³
         -(−8x⁴          -4x)
         --------------------
                +2x³     +4x
               -(2x³          +1)
               ------------------
                          4x  +1

$\color{blue}{=x^2-4x+1+\frac{4x+1}{2x^3+1}}$

(b) $\frac{x^7-128}{x-2}$

2 | 1 +0 +0 +0 +0 +0 +0 -128
       2  4  8  16 32 64 128
    ------------------------
    1  2  4  8  16 32 64  |0

$\color{blue}{=x^6+2x^5+4x^4+8^3+16^2+32x+64}$

6. Given the polynomial equation,
$x^4-2x^2-16x-15=0$

(a) list all possible rational roots

To find all possible roots, use the Rational Zero Theorem - get all of the factors of the constant ( $p$ ) and divide by the factors of the leading coefficient ( $q$ ). In the given equation, the constant is $15$ and the leading coefficient is $1$ .

$p=\pm1,\pm3,\pm5,\pm15\\q=\pm1$

$\frac{p}{q}=\pm1,\pm3,\pm5,\pm15$

The possible roots are: $\pm1,\pm3,\pm5,\pm15$

(b) find all the roots for the equation

To find the roots, plug the possible roots into the equation to test. A polynomial function will have exactly as many roots as its degree (the degree is the highest exponent of the function).

$x=1;\\1^4-2(1)^2-16(1)-15\\=1-2-16-15\\\ne0$

$x=-1;\\(-1)^4-2(-1)^2-16(-1)-15\\=1-2+16-15\\17-17\\=0$

-1 |  1   0  -2  -16  -15
         -1   1    1   15
____________________________
      1  -1  -1  -15   |0

$x^3-x^2-x-15=0$

$x=3;\\(3)^3-(3)^2-3-15\\27-9-3-15\\27-27\\=0$

3 |  1  -1  -1  -15
         3   6   15
___________________
     1   2   5   |0

$x^2+2x+5=0$

Apply Quadratic Formula:

$x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}$

$\frac{-(2)\pm\sqrt{{2}^{2}-4(1)(5)}}{2(1)}$

$=\frac{-2\pm\sqrt{-16}}{2}$

$=\frac{-2\pm4i}{2}$

$=-1\pm2i$

$x=-1, \:\:\: x=3, \:\:\: x=-1-2i, \:\:\: x=-1+2i$

7. For each of the following functions, find the domain, any vertical, horizontal, and/or slant asymptotes that may exist.

(a) $f(x)=\frac{4x^2+3}{x(x-3)}$

The denominator is $x(x-3)$ . Since any denominator must not equal zero,
the domain of the function is restricted to $x(x−3)≠0$ .

$x\ne0$
$x-3\ne0\\x\ne3$

Solving for $x$ , we get:

$x\ne0,3$

Which means, the domain is restricted to:

$x<0\:\:\:\mathrm{or}\:\:\:\:0<x<3\:\:\:\mathrm{or}\:\:\:\:x>3$

The domain in interval notation is:

$\left(-\infty \:,\:0\right)\cup \left(0,\:3\right)\cup \left(3,\:\infty \:\right)$

To find the vertical asymptotes of a rational function, set the denominator to $0$ . The numerator must be a non-zero value.

$x(x-3)=0$
$x=0,3$

Vertical asymptotes: $x=0,\:\:x=3$

To find the horizontal asymptote, let $n$ be the degree of the numerator, and $m$ the degree of the denominator.

If $n<m$ , the horizontal asymptote is the $x$ -axis (or $y=0$ )

If $n=m$ , $y=\frac{a_n}{b_m}$ is the horizontal asymptote.

If $n>m$ , the graph has a slant asymptote.

$f(x)=\frac{4x^2+3}{x(x-3)}$

Expand the denominator:

$\frac{4x^2+3}{x^2-3x}$

The degree of the numerator is $2$ . The degree of the denominator is $2$ .
$2=2$
$n=m$

Therefore, the horizontal asymptote is $y=\frac{n}{m}$

$y=\frac{4x^2}{x^2}=\frac{4}{1}=4$

The function does not have any slant asymptotes.

(b) $f(x)=\frac{x+7}{x^2+4x-21}$

Factor $x^2+4x-21$ :

$(x+7)(x-3)$

$f(x)=\frac{x+7}{(x+7)(x-3)}$

$(x+7)(x-3)\ne0$

$x+7\ne0\\x\ne-7$

$x-3\ne0\\x\ne3$

$x<-7\:\:\:\mathrm{or}\:\:\:\:-7<x<3\:\:\:\mathrm{or}\:\:\:\:x>3$

The domain of the function is:
$\left(-\infty \:,\:-7\right)\cup \left(-7,\:3\right)\cup \left(3,\:\infty \:\right)$

$\frac{x+7}{(x+7)(x-3)}$

${x+7}$ cancels out, which leaves $x-3$ .

Set the denominator to zero:

$x-3=0\\x=3$

The vertical asymptote is $x=3$

Since the degree of the numerator is less than the degree of the denominator in the function $f(x)=\frac{x+7}{x^2+4x-21} \:\:\:\:$ ( $n<m$ ), the horizontal asymptote is $y=0$ .

The function has no slant asymptote.

9. Suppose that you have $20,000 to invest.

(a) Which investment yields the greatest rate of return over 4 years: 8.25% compounded quarterly or 8.3% compounded semiannually?

$A=P(1+\frac{r}{n})^{nt}$

(b) Using the account from above, how long before you will have $40,000 in the account? Round your answer to the nearest month.

$A=P(1+\frac{r}{n})^{nt}$

$40,000=20,000(1+\frac{0.0825}{4})^{4t}$

$40,000=20,000(1.020625)^{4t}$

$\frac{40,000}{20,000}=(1.020625)^{4t}$

$2=(1.020625)^{4t}$

$\ln 2=\ln (1.020625)^{4t}$

$\ln 2=4t\ln (1.020625)$

$\frac{\ln 2}{\ln1.020625}=4t$

$4t=33.9525303983$

$t=\frac{33.9525303983}{4}$

$t= 8.48813259958\\=8.5$

8 years and 6 months

10. Using the exponential decay model, $A=A_0e^{kt}$ , find how long it will take for Xanax to decay to 90% of the original dosage, given its half-life is 36 hours. Be sure to give the function you use to find your answer.

$A=A_0e^{kt}$

$A=50$
$A_0=100$
$t=36$

$50=100e^{36t}$

Divide both sides by $100$ :

$\frac{50}{100}=e^{36k}$

Take natural log:

$\ln \frac{1}{2}=\ln e^{36k}$

$\ln \frac{1}{2}=36k$

$-0.69314718056=36k$

Divide by 36:

$k=\frac{-0.10536051565}{36}$

$k=-0.01925408834888888888888888888889$

$A=90$
$A_0=100$
$k=-0.0192541$

$90=100e^{-0.0192541t}$

$90/100=e^{-0.0192541t}$

$\ln \frac{9}{10}=-0.0192541t$

$t= \frac{\ln \frac{9}{10}}{-0.0192541}$

$t= 5.5$ hours

11. Use common logarithms or natural logarithms to rewrite the following expressions so that you can evaluate with a calculator. Round your answer to 4 decimal places.

(a) $\log_617$

$17=6^x\\\color{blue}{x=1.5812}$

(b) $\log_\pi400$

$400=\pi^x$

$\ln 400=\ln \pi^x$

$\ln 400=x \ln \pi$

$\frac{\ln 400}{\ln \pi}=x$

$\color{blue}{x=5.23395}$

12. Use the properties of logarithms to expand, or condense, each logarithmic expression as much as possible. Evaluate the logarithmic expression without a calculator when possible. Note: you don’t necessarily have to show work. However, no partial credit can be given without any work.

(a) Expand: $\log(\frac{100x^3 \sqrt[3]{3-x}}{3(x+7)^2})$

$\sqrt[3]{3-x} = (3-x)^{\frac{1}{3}}$

$3(x+7)^2 = 3(x+7)(x+7) = 3(x^2+14x+49)\\=3x^2+43x+147$

$=\log{10^2}+\log{{x}^{3}}+\log{(3-x)^{\frac{1}{3}}}-\log{(3{x}^{2}+42x+147)}$

$\color{blue}{=2\log{10}+3\log{{x}}+\frac{1}{3}\log{(3-x)}-\log{(3{x}^{2}+42x+147)}}$

(b) Condense: $\log x+ \log(x^2-1)-\log7-\log(x+1)$

Use Product Rule: $\log_b(xy)=\log_bx+\log_by$

$\log{(x\times \frac{\frac{{x}^{2}-1}{7}}{x+1})}$

$=\log{(\frac{x({x}^{2}-1)}{7(x+1)})}$

$=\log{(\frac{x(x+1)(x-1)}{7(x+1)})}$

Cancel $x+1$ :

$=\log{(\frac{x(x-1)}{7})}$

$\color{blue}{=\log{(\frac{x^2-x}{7})}}$

13. Solve the following exponential and logarithmic equations. Give the exact answer, then round your answer to 3 non-zero decimal places.

(a) $e^{4x}-3e^{2x}-18=0$

$e^{4x}-3e^{2x}=18$

$e^{4x}-e^{2x}=18/3$

$\ln e^{4x}- \ln e^{2x}= \ln 6$

$4x- 2x= \ln 6\\2x=\ln 6\\x=\frac{\ln6}{2}\\\\=0.89587973461\\\color{blue}{=0.896}$

(b) $\log(x-2)+\log5=\log100$

$\log(x-2)+\log5=\log_{10}100$

$\log(x-2)+\log5=\log_{10}10^2$

$\log(x-2)+\log5=2$

$\log_{10}{(5(x-2))}=2$

$(5(x-2))=10^2\\5x-10=100\\5x=110\\\color{blue}{x=22}$

Math

Chem

Exam 2 - Annette Storms

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