MTH 1400 (Fall 2021)
PreCalculus, Elementary Functions

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The final is comprehensive. For chapters {1,3,4,5,6} study from the 7 quizzes and TWO test that we had. For Chapter 7, go over the derivations below.
\( \S \)7.2 Sum and Difference Identities. Make sure you can derive all identities shown below.
\( \S \)7.3 Double-Angle, Half-Angle, and Reduction Formulas. Make sure you can derive all identities shown below.
\( \S \)7.4 Sum-to-Product and Product-to-Sum Formulas. Make sure you can derive all identities shown below.
\( \S \)7.5 Solving Trigonometric Equations, Problems 59-65.

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Important Identities that you should be able to derive are \begin{align} d^2_1 &= d^2_2 \\ \overbrace{\left( \cos A -\cos B\right)^2+\left( \sin A -\sin B\right)^2}^{d^2_1} &= \overbrace{\left( \cos (A-B) -1\right)^2+\left( \sin (A-B) -0 \right)^2}^{d^2_2} \\ \cos^2 A -2\cos A \cos B +\cos^2 B + \sin^2 A -2\sin A \sin B +\sin^2 B &= \cos^2 (A-B)-2\cos(A-B)+1 +\sin^2(A-B) \\ \underbrace{\cos^2 A+ \sin^2 A}_{1} -2\cos A \cos B -2\sin A \sin B +\underbrace{\sin^2 B +\cos^2 B}_{1} &= \underbrace{\sin^2(A-B)+ \cos^2 (A-B)}_{1}-2\cos(A-B)+1 \\ -2\cos A \cos B -2\sin A \sin B &= -2\cos(A-B) \\ \cos A \cos B +\sin A \sin B &= \cos(A-B) \,\,\, \checkmark \end{align}

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\begin{align} \cos(\alpha \pm\beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\ \sin(\alpha \pm\beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \end{align} Comparing distances of cords on unit circle, we started by deriving the big mama identity $$ \cos(\alpha -\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \tag{Big Mama} $$ Then we used the substitution \( \boxed{\cos(\alpha +\beta) = \cos(\alpha -(-\beta))} \) and the fact(s) that $$ \underbrace{\cos(-x)=\cos(x)}_{\text{even}} \hspace{3mm} \text{ and } \hspace{3mm} \underbrace{\sin(-x)=-\sin(x)}_{\text{odd}} $$ to derive the older daughter identity $$ \cos(\alpha +\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $$ Next order of business was to apply \(\boxed{ \sin (x)=\cos\left(x-\frac{\pi}{2} \right) }\) to the older daughter identity to get \begin{align} \sin (\alpha+ \beta) &= \cos\left((\alpha+\beta)-\frac{\pi}{2} \right) \\ &= \cos\left( \alpha+\left(\beta-\frac{\pi}{2}\right) \right) \\ &= \cos \alpha \, \underbrace{\cos\left( \beta-\frac{\pi}{2} \right)}_{\sin \beta}- \sin \alpha \, \underbrace{\sin\left( \beta-\frac{\pi}{2} \right)}_{-\cos \beta} \\ &= \cos \alpha \sin \beta +\sin \alpha \cos \beta \end{align} and finally \begin{align} \sin (\alpha- \beta) &= \sin \left(\alpha+ (-\beta)\right) \\ &= \cos \alpha\, \underbrace{\sin (-\beta)}_{-\sin\beta} +\sin \alpha\, \underbrace{\cos (-\beta)}_{\cos \beta} \\ &= \sin \alpha \cos \beta -\cos \alpha \sin\beta \end{align} A very special case(s) of colosal importance are the so called double angle identities \begin{align} \sin 2\phi &= \sin\left(\phi+\phi \right)\\ &= \cos \phi \sin \phi +\cos\phi \sin \phi \\ &= 2\cos\phi \sin\phi \\ \cos 2\theta &= \cos\left(\theta+\theta \right) \\ &= \cos\theta \cos\theta -\sin\theta \sin\theta \\ &= \cos^2\theta -\sin^2\theta = \underbrace{1-\sin^2\theta}_{\cos^2\theta}-\sin^2\theta = 1-2\sin^2\theta \\ &= \cos^2\theta-\underbrace{\left( 1-\cos^2\theta \right)}_{\sin^2\theta} \\ &= 2\cos^2\theta -1 \end{align} The last two identities were obtained by solving \(\boxed{\cos 2\theta = 2\cos^2\theta -1}\) for \(\cos^2\theta\) and, \(\boxed{\cos 2\theta = 1-2\sin^2\theta}\) for \(\sin^2\theta\), and they are $$ \boxed{\cos^2\theta =\frac{1}{2}\left[ 1+\cos 2\theta \right]} \hspace{3mm} \text{ and } \hspace{3mm} \boxed{\sin^2\theta =\frac{1}{2}\left[ 1-\cos 2\theta \right]} $$ By substituting (aka change of variable) \(\boxed{x=2\theta}\) which is equivalent to \(\boxed{\theta=\frac{x}{2}}\) we get less important half angle identities $$ \boxed{\cos^2\frac{x}{2} =\frac{1}{2}\left[ 1+\cos x \right]} \hspace{3mm} \text{ and } \hspace{3mm} \boxed{\sin^2\frac{x}{2} =\frac{1}{2}\left[ 1-\cos x \right]} $$

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We can obtain one more form from big mama and her daughters. The so called product identities. The big mama and her daughters are enumerated here for easy reference \begin{align} \cos(\alpha +\beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \tag{1} \\ \cos(\alpha -\beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \tag{2} \\ \sin(\alpha +\beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \tag{3} \\ \sin(\alpha -\beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \tag{4} \end{align} Add (3) to (4), then divide by 2 to get $$ \sin \alpha \cos \beta = \frac{1}{2}\left[ \sin(\alpha +\beta) + \sin(\alpha -\beta) \right] \tag{5} $$ Add (1) to (2), then divide by 2 to get $$ \cos \alpha \cos \beta = \frac{1}{2}\left[ \cos(\alpha-\beta)+\cos(\alpha+\beta)\ \right] \tag{6} $$ Finally, \(\frac{(2)-(1)}{2}\) gives $$ \sin\alpha\sin\beta = \frac{1}{2}\left[ \cos(\alpha-\beta)-\cos(\alpha+\beta) \right] \tag{7} $$

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Use a change of variable \(\alpha = \frac{u+v}{2}\) and \(\beta = \frac{u-v}{2}\) to obtain sum-to-product identities. The figure above, illustrates the identity $$ \sin A + \sin B = 2 \cdot \overbrace{\sin \left( \frac{A+B}{2} \right)}^{\text{average period}} \cdot \overbrace{\cos \left( \frac{A-B}{2} \right)}^{\text{slow envelope}} $$

• Had Quiz 5
• \(\S\)4.3 Logarithmic functions.
  The core property: $$ \boxed{ f^{-1}(x)=\log_b(x) } \,\,\,\,\, \text{ is the inverse of } \,\,\,\,\, \boxed{f(x)=b^x}$$ It is very important to get used to the notation \begin{align} \boxed{y=\log_b x} &\iff \boxed{x=b^y} \\ \boxed{2=\log_7 49} &\iff \boxed{49=7^2} \\ \boxed{-\frac{3}{2}=\log_4 \frac{1}{8}} &\iff \boxed{\frac{1}{8}=4^{-\frac{3}{2}}} \\ \boxed{-\frac{2}{3}=\log_\frac{1}{8} 4} &\iff \boxed{4=\left(\frac{1}{8} \right)^{-\frac{2}{3}}} \end{align} Moreover we have \begin{align} f^{-1}\left(f(x) \right) &= \log_b\left( b^x \right)=x \\ f\left(f^{-1}(x) \right) &= b^{\log_b x}=x \end{align}
• \(\S\)4.4 Graphs Logarithmic functions.
• \(\S\)4.5 Logarithmic Properties
  Let $$ \boxed{\boxed{y_1 = \log_b M} \iff \boxed{M=b^{y_1}}} \,\,\,\,\, \text{ and } \,\,\,\,\, \boxed{\boxed{y_2 = \log_b N} \iff \boxed{N=b^{y_2}}} $$ Then it follows

  (start with) Exponential form		Logarithmic form
  \( M\cdot N=b^{y_1}\cdot b^{y_2}=b^{y_1+y_2} \)	\(\iff\)	\(\log_b(M\cdot N)=\underbrace{\log_b M + \log_b N}_{y_1+y_2}\)	Product property
  \( \frac{M}{N}=\frac{b^{y_1}}{b^{y_2}}=b^{y_1-y_2} \)	\(\iff\)	\(\log_b\left(\frac{M}{N} \right)=\underbrace{\log_b M - \log_b N}_{y_1-y_2}\)	Quotient property
  \( N^{y_1}=\left(b^{y_2} \right)^{y_1}=b^{y_2\cdot y_1} \)	\(\iff\)	\(\log_b\left(N^{y_1} \right)=\underbrace{y_1\cdot \log_b N}_{y_1\cdot y_2}\)	Power property
  \( \begin{align} b^{y_1} &= M \\ \log_a \left( b^{y_1}\right) &= \log_a M \\ y_1\cdot \log_a b &= \log_a M \\ y_1 &= \frac{\log_a M}{\log_a b} \end{align} \)	\(\iff\)	\( \log_b M=\frac{\log_a M}{\log_a b}\)	Change of Base

• \(\S\)4.6 Exponential and Logarithmic Equations

1. Had Quiz 4 (complex roots, conjugates, and long division).
2. Finished \(\S3.7\) Rational Functions.
3. Started Chapter 4 (Exponential and Logarithmic Functions), \(\S4.1\) and \(\S4.2\)
   1. Properties of Exponential Functions.
   2. Graphs of Exponential Functions.
   3. Compound interest $$ A(t)=P_0\left(1+\frac{r}{k} \right)^{kt} $$ Here: \(A(t)\) is the amount in the account in \(\$\), at time \(t\) years. \(P_0\) is the principal amount that is deposited initially at the account. \(r\) is the annual interest rate. \(k\) number of times compounding happens in a year.
   4. For continuous compounding, we need to:
      • Send \(k\rightarrow \infty\)
      • Apply \(\boxed{\lim\limits_{x\rightarrow \infty} \left(1+\frac{1}{x} \right)^x=e}\) to \(\boxed{A(t)=P_0\left(1+\frac{r}{k} \right)^{kt}}\)
        
        
        
        
      • Template match to identify \(\boxed{\frac{1}{x}=\frac{r}{k}}\). Hence \(\boxed{k=rx}\) and more importantly: as \(\boxed{x \rightarrow \infty}\) so does \(\boxed{k \rightarrow \infty}\) $$ \begin{align} \lim_{k\rightarrow \infty} P_0 \left(1+\frac{r}{k} \right)^{kt} &= \lim_{x\rightarrow \infty} P_0 \left(1+\frac{1}{x}\right)^{rxt} \\ &= \lim_{x\rightarrow \infty} P_0 \left[\left(1+\frac{1}{x}\right)^x\right]^{rt} \\ &= P_0 \left[\lim_{x\rightarrow \infty} \left(1+\frac{1}{x}\right)^x\right]^{rt} \\ &= P_0 e^{rt} \end{align} $$

1. problems 27-37 using LONG DIVISION
2. problems 42, 43, 51 and 53 use LONG DIVISION ONLY

Fundamental Theorem of Algebra

1. Every odd degree polynomial has at least one real zero. So you can plot on calculator and see the real zero.
2. However even degree polynomials are not guaranteed to have real zeros.
3. If a polynomial with real coefficients has a complex zero \(x_1=a+bi\), then it must have another zero \(x_2=a-bi\), complex conjugates.

Problems for quiz 4 October 19, Tuesday

1. Use a graphing calculator to find a real zero of \(f(x)\).
2. Find all the zeros (aka roots) of \(f(x)\).
3. Write \(f(x)\) as a product of linear factors.
4. Verify the correctness of your factorization above by direct multiplication.

• The \(\mathbb{R}\)eal (potentially \(\mathbb{C}\)omplex) numbers \(a_0,a_1,a_2,\ldots , a_{n-1},a_n\) are called coefficients.
• Each individual term \(a_k x^k\) is called a monomial.
• The monomial \(a_nx^n\) is called the leading term.
• The coefficient \(a_n\) is called the leading coefficient.
• The coefficient \(a_0\) is called the constant coefficient.

• The leading coefficient is \(a_4=-\frac{3}{2}\).
• The constant coefficient is \(a_0=0\).

• \(\boxed{x=-4}\) with multiplicity THREE, hence \(f(x)\) has a factor \((x+4)^3\).
• \(\boxed{x=1}\) with multiplicity ONE, hence \(f(x)\) has a factor \((x-1)\).
• \(\boxed{x=3}\) with multiplicity TWO, hence \(f(x)\) has a factor \((x-3)^2\).
• \(y\)-intercept \((0,-4)\).

Quadratic Optimization Problems due September 30.

In Quadratic Optimization Problems it is absolutely mandatory to:

1. Declare the variables clearly. (not more than TWO in our problems).
2. State the objective to be optimized (minimized or maximized).
3. State the constraints when necessary.
4. Find the optimum value of the objective, that is the vertex of the quadratic function.
5. Write down your conclusion in a clear language using complete sentences.

1. Difference Quotients: Simplify as much as possible (till there is no division by \(\boxed{h=0}\) the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for each of the following: $$ \begin{align} f(x) &= 3x^2-5x+7 \\ g(x) &= \frac{-3}{x+5} \\ u(x) &= \sqrt{x+7} \\ w(x) &= \frac{1}{\sqrt{x+7}} \end{align} $$
2. Transformation of functions: Sketch without calculators, and identify the vertex and all the intercepts of: $$ \begin{align} f_1(x) &= -2|x+4|+7 \\ f_2(x) &= -3|x-5|-1 \\ f_3(x) &= +4|x+1| \\ f_3(x) &= +4|x+1|+3 \end{align} $$
3. Inequality: Solve: $$ \begin{align} 2|3x+1| & < 5 \\ 2|3x-1| & \geq 5 \\ -2|3x-2|+8 & \leq 1 \\ -2|3x-2|+8 & > 11 \end{align} $$
4. Complex variables: Let \(z= \left( \frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}} \right) \), \(z_1=-3+5i\), \(z_2=5-2i\) and \(z_3=-1+2i\), simplify each of the following: $$ \frac{z_1-z_2}{z_3}, \,\, \frac{z_2}{z_1+z_3}, \,\, z^2, \,\,\, zz^*, \,\, (z^*)^2 $$
5. Parabolas in vertex form: Sketch without calculators, and identify the vertex, the axes of symmetry, and all the intercepts (state the complex valued solutions in case if there are no real \(x\)-intecepts): $$ \begin{align} f_1(x) &= -2(x+4)^2+7 \\ f_2(x) &= -3(x-5)^2-1 \\ f_3(x) &= +4(x+1)^2 \\ f_3(x) &= +4(x+1)^2+3 \end{align} $$

• Absolute value inequalities have two main forms:
  • \(\boxed{|x|< 7}\) with solution set \( -7< x < 7 \), or \((-7,7)\), or \(]-7,7[\) a single connected finite length interval.
  • \(\boxed{|x|\geq 3}\) with solution set \( (-\infty,-3]\cup [+3,\infty) \), or \( \{ x\in\mathbb{R}\, \mathbf{|} \, x\leq -3 \,\, \text{ or } \,\, x\geq +3 \} \), two disjoint infinite length intervals.
• Complex numbers \(z=a+bi\) where \(\boxed{i^2=-1}\) or \(\boxed{ i=+\sqrt{-1}} \)
  Let \(z_1=a+bi\) and \(z_2=c+di\), then we define
  • Addition: \(z_1+z_2=(a+bi)+(c+di)=\underbrace{(a+c)}_{\text{real part}}+\underbrace{(b+d)}_{\text{imaginary}}i\)
  • Subtraction: \(z_1-z_2=(a+bi)-(c+di)=\underbrace{(a-c)}_{\text{real part}}+\underbrace{(b-d)}_{\text{imaginary}}i\)
  • Multiplication: \(z_1\cdot z_2=(a+bi)(c+di)=ac+adi+bci+bdi^2=\underbrace{(ac-bd)}_{\text{real}}+\underbrace{(ad+bc)}_{\text{imaginary}}i \)
  • Conjugate of \(\boxed{z=a+bi}\) is \(\boxed{z^*=a-bi}\), and note the following: $$ \begin{align} \frac{z+z^*}{2} &= \frac{(a+bi)+(a-bi)}{2} = a = \mathcal{Re}(z) \\ \frac{z-z^*}{2i} &= \frac{(a+bi)-(a-bi)}{2i} = b = \mathcal{Im}(z) \\ z\cdot z^* &= (a+bi)(a-bi)=\underbrace{a^2+b^2 = |z|^2 =\,\,\, \text{hypotenuse}^2}_{\text{Pythagorean Theorem}} \end{align} $$
  • Division: (multiply both numerator and denominator by the conjugate of the denominator) $$ \begin{align} \frac{z_1}{z_2} &= \frac{a+bi}{c+di} \\ &= \frac{a+bi}{c+di} \cdot \frac{\color{red}{c-di}}{\color{red}{c-di}} \\ &= \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} \\ &= \underbrace{\frac{ac+bd}{c^2+d^2}}_{\text{real part of }\frac{z_1}{z_2}}+\underbrace{\frac{bc-ad}{c^2+d^2}}_{\text{imaginary part of }\frac{z_1}{z_2}}\cdot i \end{align} $$
  • Powers of \(i\). $$ i^1 = i, \,\,\, i^2=-1,\,\,\, i^3=-i,\,\,\, i^4=1,\,\,\, i^5=i,\,\,\, i^6=-1,\,\,\, i^7=-i,\,\,\, i^8=1,...\,\,\, i^{37}=i,... \,\,\, i^{63}=-i $$ and as importantly $$ i^{-1}=\frac{1}{i}=\frac{i\cdot i \cdot i \cdot i}{i}=\frac{i^3}{1}=-i $$
• Parabolas: can have two forms $$ \underbrace{f(x)=ax^2+bx+c}_{\text{standard form}} \hspace{8mm} \text{ and } \hspace{8mm} \underbrace{f(x)=a(x-k)^2+h}_{\text{vertex form}} $$ We did two graphing examples:
  1. Graph \( f(x)=-2(x+3)^2+5 \), and find the vertex, the intercepts and the axis of symmetry.
     Solution: The parent function \(y=x^2\) is
     1. Shifted left 3 units to get \( y=(x+3)^2 \)
     2. Stretched by 2 and reflected about \(x\)-axis to get \( y=-2(x+3)^2 \)
     3. Shifted up 5 units to get \( f(x)=-2(x+3)^2+5 \)
     The vertex is \( (-3,5) \). Axis of symmetry is \(\boxed{x=-3}\). For \(y\)-intercept set \(x=0\) to obtain \((0,f(0))=(0,-13)\).
     For \(x\)-intercept set \(y=0\) and solve for \(x\). This requires more steps: $$ \begin{align} -2(x+3)^2+5 &= 0 \\ (x+3)^2 &= \frac{5}{2} \\ x+3 &= \pm\sqrt{\frac{5}{2}} \\ x &= -3\pm\sqrt{\frac{5}{2}} \end{align} $$ Hence the \(x\)-intercepts are \(\left(-3-\sqrt{\frac{5}{2}},0\right)\) and \(\left(-3+\sqrt{\frac{5}{2}},0\right)\)
     
  
  

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  2. Graph \( g(x)=\frac{1}{3}(x-2)^2+5 \), and find the vertex, the intercepts and the axis of symmetry.
     Solution: The parent function \(y=x^2\) is
     1. Shifted right 2 units to get \( y=(x-2)^2 \)
     2. Squeezed by \(\frac{1}{3}\) to get \( y=\frac{1}{3}(x-2)^2 \)
     3. Shifted up 5 units to get \( g(x)=\frac{1}{3}(x-2)^2+5 \)
     The vertex is \( (2,5) \). Axis of symmetry is \(\boxed{x=2}\). For \(y\)-intercept set \(x=0\) to obtain \((0,g(0))=(0,\frac{4}{3}+5)\).
     For \(x\)-intercept set \(y=0\) and solve for \(x\). This requires more steps: $$ \begin{align} \frac{1}{3}(x-2)^2+5 &= 0 \\ (x-2)^2 &= -15 \\ x-2 &= \pm\sqrt{-15}=\pm\sqrt{(15)(-1)}=\pm\sqrt{15}\cdot \underbrace{\sqrt{-1}}_{i}=\pm\sqrt{15}\cdot i=\pm i\cdot \sqrt{15} \\ x &= 2\pm\sqrt{15}\, i \end{align} $$ Hence this parabola does not have real \(x\)-intercepts.

• Difference Quotients: \(\frac{\text{rise}}{\text{run}}=\frac{f(x+h)-f(x)}{h}=\frac{f(x)-f(a)}{x-a}\)
  It is important here to simplify these quotients as much as possible till there is no division by zero when substituting \(\boxed{h=0}\) or \(\boxed{x=a}\). The four main examples you are expected to simplify on the next quiz are the difference quotients for: $$ \begin{align} f(x) &= 3x^2-5x+7 \\ g(x) &= \frac{-3}{x+5} \\ u(x) &= \sqrt{x+7} \\ w(x) &= \frac{1}{\sqrt{x+7}} \end{align} $$
  
  Here is the Matlab code i wrote to generate this gif.
  

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• Transformations of Functions: given a basic function such as $$\left\{w(x)=\frac{1}{x},\,\, h(x)=\sqrt{x},\,\, g(x)=x^2, \,\, k(x)=x^3, \,\, m(x)=|x|, \,\, w(x)=\sqrt[3]{x}\right\}$$
  1. The graph of \(y=f(x-a)\) is a horizontal translation of the graph of \(y=f(x)\), or delay by \(a\), i.e shift right if \(a>0\).
  2. The graph of \(y=f(-x)\) is a reflection of the graph of \(y=f(x)\) about \(y\)-axis. Some functions (called EVEN) are not affected by this, like \(g(x)=x^2\) or \(m(x)=|x|\).
  3. The graph of \(y=-f(x)\) is a reflection of the graph of \(y=f(x)\) about \(x\)-axis.
  4. The graph of \(y=f(x)+a\) is a vertical translation of the graph of \(y=f(x)\), up by \(a\) if \(a>0\).
  5. The graph of \(y=af(x)\) is a vertical stretch of the graph of \(y=f(x)\) if \(1< a\), and a vertical squeeze of \(y=f(x)\) if \(0< a <1 \).
  6. The graph of \(y=f(ax)\) is a horizontal squeeze of the graph of \(y=f(x)\) if \(1< a\), and a horizontal stretch of \(y=f(x)\) if \(0< a <1 \).

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  As a quick exercise, sketch the graphs of the following without using calculator, and find the \(x\)-intercepts, the vertex and the \(y\)-intercept: $$ \begin{align} f_1(x) &= -2|x+4|+7 \\ f_2(x) &= -3|x-5|-1 \\ f_3(x) &= +4|x+1| \\ f_4(x) &= -2(x+4)^2+7 \\ f_5(x) &= -3(x-5)^2-1 \\ f_6(x) &= +4(x+1)^2 \end{align} $$

1. $$\boxed{ \begin{align} (f \circ g)(3) &= f(g(3)) \\ &= f\left( 3^2+1 \right) \\ &= f(10) \\ &= \frac{1}{10+1} = \frac{1}{11} \end{align} } $$
2. $$ \boxed{ \begin{align} (g \circ f)(3) &= g(f(3)) \\ &= g\left( \frac{1}{3+1} \right) \\ &= g\left( \frac{1}{4} \right) \\ &= \left( \frac{1}{4} \right)^2+1 \\ &= \frac{1}{16}+1=\frac{17}{16} \end{align} } $$
3. $$ \boxed{ \begin{align} (f \circ g)(z+1) &= f(g(z+1)) \\ &= f\left( (z+1)^2+1 \right) \\ &= f\left( z^2+2z+2 \right) \\ &= \frac{1}{[z^2+2z+2]+1} \\ &= \frac{1}{z^2+2z+3} \end{align} } $$

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If \(f\) is a one-to-one, that is: \(f(x_1)\neq f(x_2)\) for distinct \(x_1\neq x_2\), then the unique inverse function \(f^{-1}\) exists and has the property $$\begin{align} f^{-1}(f(x)) &= (f^{-1}\circ f)(x) \\ &= I(x) \\ &= x \\ f(f^{-1}(x)) &= (f\circ f^{-1})(x) \\ &= I(x) \\ &= x \end{align} $$ To find \(f^{-1}(x)\) we use the following 4 steps:
1. Write \(y=f(x)\)
2. Swap \(x\) and \(y\) to get \(x=f(y)\)
3. Solve for \(y\)
4. Name the result in 3. above as \(y=f^{-1}(x)\)
Couple of examples: Example 1 and Example 2.

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Basic Toolkit Functions include:
1. Lines \(\boxed{f(x)=mx+b}\), one-to-one if \(m\neq 0\), not one-to-one if \(m=0\).
2. Absolute value function \(\boxed{f(x)= |x|}\), not one-to-one.
3. Parabola \(\boxed{f(x)=x^2}\), not one-to-one.
4. Square root \(\boxed{f(x)= \sqrt{x}}\), one-to-one.
5. Cubic function \(\boxed{f(x)= x^3}\), one-to-one.
6. Cube root \(\boxed{f(x)= \sqrt[3]{x}}\), one-to-one.
7. Reciprocal function \(\boxed{f(x)= \frac{1}{x}}\), one-to-one.
8. Reciprocal squared \(\boxed{f(x)= \frac{1}{x^2}}\), not one-to-one.



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Problems:
• \(\S\)1.1: 12-26, 28-30, 32, 33, 37-39, 40-59, 74, 75.
• \(\S\)1.2: even 8-24, 28-36, 40-44.
• \(\S\)1.3: even 6-14, all 18-25.
• \(\S\)1.4: even 6-10, 11, even 12-16, all 20-25, even 34-40, all 42-49, 50-57.
• \(\S\)1.5: 18-23, 27-30, 33-46, 47-52, 69-77, 78, 81.
• \(\S\)1.6: even 16-36, even 38-52.
• \(\S\)1.7: 11, 12, 13-15, 17, 18, 23-32.