If $x>1$, $x^2$ is always greater than $dfrac{1}{x^2}$

For example when $x=2$

$x^2=4$ and $x^{-2}=dfrac{1}{4}$

A number divided by something that has absolute value of less than 1 will be greater than the original number.

For example when $x=dfrac{1}{2}$

$x^{-2}=dfrac{1}{x^2}=dfrac{1}{(1/2)^2}=4$

$x^2=left(dfrac{1}{2}right)^2=dfrac{1}{4}$

b) $dfrac{1}{x^2}>x^2$ when $|x|<1$ and $xneq 0$

$$
=(-7)^{-1}=dfrac{1}{-7}=-dfrac{1}{7}
$$

$x^acdot x^b=x^{a+b}$

$$
x^{-a}=dfrac{1}{x^a}
$$

$(-x)^n$ = $-x^n$ if $n$ is odd

$(-x)^n=x^n$ if $n$ is even

for $n>1$

$$
=dfrac{5^3}{5^3}=5^{3-3}=5^0=1
$$

$dfrac{x^a}{x^b}=x^{a-b}$

$=dfrac{4^{-10}(4^{-18})}{(4^{-32})}$

$=dfrac{4^{-10+(-18)}}{4^{-32}}$

$=4^{-28-(-32)}$

$=4^{-28+32}$

$$
=4^4=256
$$

$x^acdot x^b=x^{ab}$

$dfrac{x^a}{x^b}=x^{a-b}$

$$
=11^{9+(-7)}=11^2=121
$$

$$
x^acdot x^b=x^{a+b}
$$

$$
=left(dfrac{(-3)^{7+4}}{(-3)^{12}} right)^{-3}
$$

$=left( dfrac{(-3)^{11}}{(-3)^{12}}right)^{-3}$

$=left((-3)^{11-12}right)^{-3}$

$=left((-3)^{-1}right)^{-3}$

$=(-3)^{(-1)(-3)}$

$=(-3)^3=-3^3$

$$
=-27
$$

$dfrac{x^a}{x^b}=x^{a-b}$

$(x^a)^b=x^{ab}$

$(-x)^n$ = $-x^n$ if $n$ is odd

$(-x)^n=x^n$ if $n$ is even

for $n>1$

b) $(-2)^5=-32$

c) $5^0=1$

d) $4^4=256$

e) $11^2=121$

f) $(-3)^3=-27$

a) $sqrt[3]{x^7}=x^{frac{7}{3}}$

b) $y^{frac{8}{5}}=sqrt[5]{y^8}$

c) $(sqrt{p})^{11}=(p^{frac{1}{2}})^{11}=p^{frac{11}{2}}$

d) $m^{1.25}=m^{1+frac{1}{4}}=m^{frac{5}{4}}=sqrt[4]{m^5}$

b) $sqrt[5]{y^8}$

c) $p^{frac{11}{2}}$

d) $sqrt[4]{m^5}$

$left(dfrac{x}{y}right)^{-a}=left(dfrac{y}{x}right)^a=dfrac{y^a}{x^a}$

$sqrt[m]{a^n}=a^{frac{n}{m}}$

$(x^a)^b=x^{ab}$

$$
sqrt[m]{a^m}=a
$$

b) $dfrac{15}{4}$

c) $-dfrac{1}{15}$

d) $81$

e) $-64$

$$
(x^a)(x^b)=x^{a+b}
$$

$=dfrac{c^{1+frac{5}{6}}}{c^2}$

$=dfrac{c^{frac{11}{6}}}{c^2}$

$=c^{frac{11}{6}-2}$

$=c^{frac{11}{6}-frac{12}{6}}$

$=c^{-frac{1}{6}}$

$$
=dfrac{1}{c^{frac{1}{6}}}
$$

$dfrac{x^a}{x^b}=x^{a-b}$

$$
x^{-a}=dfrac{1}{x^a}
$$

$(x^a)^b=x^{ab}$

$dfrac{x^a}{x^b}=x^{a-b}$

b) $b^1$

c) $dfrac{1}{c^{frac{1}{6}}}$

d) $d^{frac{13}{2}}$

e) $e^{14}$

f) $f^{frac{1}{5}}$

Let $a=5^2$ and $b=6^2$

$sqrt{5^2+6^2}=sqrt{25+36}=sqrt{61}approx 7.81$

$sqrt{a}+sqrt{b}=sqrt{5^2}+sqrt{6^2}=5+6=11$

This proves that $sqrt{a+b} neq sqrt{a}+sqrt{b}$

$sqrt{5^2+6^2}=sqrt{25+36}=sqrt{61}approx 7.81$

$sqrt{a}+sqrt{b}=sqrt{5^2}+sqrt{6^2}=5+6=11$

This proves that $sqrt{a+b} neq sqrt{a}+sqrt{b}$

$(xy)^a=x^ay^a$

$x^acdot x^b=x^{a+b}$

$(-x)^a=x^a$ if $a$ is even

$(-x)^a=-x^a$ if $a$ is odd

$dfrac{x^a}{x^b}=x^{a-b}$

$x^acdot x^b=x^{a+b}$

$dfrac{x^a}{x^b}=x^{a-b}$

$dfrac{x^a}{x^b}=x^{a-b}$

$(x^a)^b=x^{ab}$

$x^acdot x^b=x^{a+b}$

$dfrac{x^a}{x^b}=x^{a-b}$

$(-x)^a=x^a$ if $a$ is even

$(-x)^a=-x^a$ if $a$ is odd

b) $dfrac{64}{m^2}=4$

c) $dfrac{3}{2w^3}=-dfrac{1}{18}$

d) $3y^5=-96$

e) $dfrac{1}{6}x^{12}=dfrac{2048}{3}$

f) $dfrac{16}{3x}=dfrac{32}{3}$

$sqrt[m]{a^n}=a^{frac{n}{m}}$

$(x^a)^b=x^{ab}$

$sqrt{a}=a^{frac{1}{2}}$

$(x^a)^b=x^{ab}$

$$
x^{-a}=dfrac{1}{x^a}
$$

$(x^a)^b=x^{ab}$

$$
x^{-a}=dfrac{1}{x^a}
$$

$x^acdot x^b=x^{a+b}$

$sqrt[m]{x^n}=x^{frac{n}{m}}$

$dfrac{x^a}{x^b}=x^{a-b}$

$(-x)^a=-x^a$ if $a$ is odd

$(x^a)^b=x^{ab}$

$x^{-a}=dfrac{1}{x^a}$

$(-x)^{frac{a}{b}} = x^{frac{a}{b}}$ if $b$ is odd and $a$ is even

$(-x)^{frac{a}{b}} = -x^{frac{a}{b}}$ if both $a$ and $b$ are odd

$(-x)^{frac{a}{b}}=text{undefined}$ if $b$ is even

$sqrt[m]{x^n}=x^{frac{n}{m}}$

$dfrac{x^a}{x^b}=x^{a-b}$

b) $dfrac{b}{a}$

c) $dfrac{1}{m^2n^{frac{1}{2}}}$

d) $x^9$

e) $-dfrac{1}{x^{1.8}}$

f) $dfrac{x^9}{y}$

a) The second differences are constant so it is quadratic.\\

exponential function $implies$ has an asymptote

quadratic function $implies$ parabolic and symmetric

linear $implies$ straight line

a) The graph has an asymptote so it must be exponential.

b) The graph is parabolic so it must be quadratic.

c) The graph has an asymptote so it must be exponential.

(1) horizontal stretching by a factor of $2$ $implies left(dfrac{1}{2}right)^{frac{x}{2}}$

(2) vertical translation 3 units down $implies left(dfrac{1}{2}right)^{frac{x}{2}}-3$

(1) vertical stretching by a factor of $dfrac{1}{4}implies dfrac{1}{4}(2^x)$

(2) reflection in the $y$-axis $implies dfrac{1}{4}(2^{-x})$

(3) vertical translation 1 unit upwards $implies dfrac{1}{4}(2^{-x})+1$

(1) reflection in $x$-axis $implies -3^{x}$

(2) vertical stretching by a factor of 2 $implies -2(3^x)$

(3) horizontal compression by a factor of 2 $implies -2(3^{2x})$

(4) horizontal translation 2 units to the left $implies -2(3^{2(x+2)})=-2(3^{2x+4})$

(1) reflection in the $x$-axis $implies$ $-5^x$

(2) vertical compression by a factor of $dfrac{1}{10} implies -dfrac{1}{10}(5^x)$

(3) horizontal compression by a factor of 3 $implies =-dfrac{1}{10}left(5^{3x}right)$

(4) horizontal translation 3 units to the right $implies -dfrac{1}{10}left(5^{3(x-3)}right)$

(5) vertical translation 10 units upwards $implies -dfrac{1}{10}left(5^{3(x-3)}right)+10$

$y$-intercept: value of $y$ when $x=0$

asymptote: $y=c$

$y$-intercept = 2

By inspection, we can see that the asymptote is $y=1$

Thus, this equation must be of the form

$y=acdot b^{k(x-d)}+1$

Since there are no vertical and horizontal scaling mentioned, $a=1$ and $k=1$

Since there is no horizontal translation, $d=0$

The equation is therefore of the form
$y=b^{x}+1$

We see that the graph passes through $(1,3)$ so we can solve for $b$ as

$3=b^{1}+1implies b=3-1=2$

The equation is therefore $y=2^x+1$

asymptote: $y=1$

equation: $y=2^x+1$

$a=$ the initial amount or number

growth or decay rate = $b-1$

growth if $b>1$

decay if $b<1$

base $>1implies$ growth

base $<1implies$ decay

In this case, base = $frac{1}{2}<1$, thus it is an exponential decay.

$T(t)=69left(dfrac{1}{2}right)^{frac{t}{30}}+21$

$T(0)=69left(dfrac{1}{2}right)^{frac{0}{30}}+21$

$T(0)=69(1)+21=90$

The initial temperature of the coffee is $90^circ$ C

Perform the following transformations.

(1) horizontal stretching by a factor of 30 $implies y=left(dfrac{1}{2}right)^{frac{t}{30}}$

(2) vertical stretching by a factor of 69 $implies y=69left(dfrac{1}{2}right)^{frac{t}{30}}$

(3) vertical translation 21 units upward $implies y=69left(dfrac{1}{2}right)^{frac{t}{30}}+21$

$T(48)=69left(dfrac{1}{2}right)^{frac{48}{30}}+21=44^circ$ C

b) $90^circ$ C

c) graph has been plotted in the answers

d) 44 $^circ$ C

e) The 30 in the exponent must be lesser.

f) horizontally compressed

$V(n)=28;000(0.875)^n$

$V(0)=28;000$

$bold{b)}$ The decay rate of $y=ab^x$ is $b-1$. In this case,

$0.875-1=-0.125$

The annual rate of depreciation is $12.5%$

$bold{c)}$ The value at the end of 3 years is the value of V when $n=3$

$V(3)=28;000(0.875)^3=$18758$

$bold{d)}$ 30 months = $dfrac{30}{12}=2.5$ years

$V(2.5)=28;000(0.875)^{2.5}=$20;053$

$bold{e)}$ The change in value in the first year is $V(1)-V(0)$

$V(0)-V(1)=28000(0.875)-28000=-$3500$

It lost $$3500$ in the first year.

$bold{f)}$ The value lost in the fifth year is $V(5)-V(4)$

$V(5)-V(4)=28000(0.875)^5-28000(0.875)^4=-2051.64$

It lost $$2052$ in the fifth year

b) $12.5%$

c) $$18;758$

d) $$20;053$

e) $$3500$

f) $$2052$

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the quantity after $n$ periods

$A_0$ is the initial quantity

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$

a) Let $P$ be the fraction of the pond covered by water lily and $n$ be the number of weeks.

“each week they increase by 10$%$” $implies r=0.10$

“cover one-third of a pond now” $implies P_0=frac{1}{3}$

$P=P_0(1+r)^n$

$P=dfrac{1}{3}(1+0.10)^n$

$P=dfrac{1}{3}(1.1)^n$

Each part of the equation represents the following:

$dfrac{1}{3}$ signifies that initially, $dfrac{1}{3}$ of the pond is covered by lilies

$1.1$ signifies the 10$%$ increase in coverage each week

$n$ is the number of weeks elapsed

“half-life of $4.5times 10^9$ years” $implies r=-dfrac{1}{2}$ and $n=dfrac{t}{4.5times 10^9}$

$A=A_0left(1-dfrac{1}{2}right)^{frac{t}{4.5times 10^9}}$

$A=A_0left(dfrac{1}{2}right)^{frac{t}{4.5times 10^9}}$

Each part of the equation represents the following:

$A_0$ is the initial amount of $U_{238}$

$dfrac{1}{2}$ signifies that the amount is halved every certain period

$4.5times 10^9$ signifies the time at which it will be halved

$t$ is the number of years.

“each gel reduces the intensity by $4%$” $implies r=-0.04$

Assume initial value of $100%$

$I=100(1-0.04)^n$

$I=100(0.96)^n$

Each part of the equation represents the following.

$0.96$ signifies that it decreases by $4%$ per gel

$n$ is the number of gels

b) $A=A_0left(dfrac{1}{2}right)^{frac{t}{4.5times10^9}}$

c) $I=100(0.96)^n$

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$

a) Let $P$ be the population at any year, $P_0$ be the population in the year $1990$, and $n$ be the number of years after 1990.

“growing at a rate of 3$%$ per year” $implies r=0.03$

“In 1990, the population was 45 000” $implies P_0=45;000$

$P=P_0(1+r)^n$

$P=45;000(1+0.03)^n$

$P=45;000(1.03)^n$

In this case, equation

$P$ is the population at any $n$

$45,000$ is the population in 1990

$1.03$ signifies the 3$%$ increase per year

$n$ is the number of years after 1990

$P=45;000(1.03)^{17}=74;378$

c) The population doubles when $P=2times 45;000=90;000$

By graphical approach, we see that it would be 24 years after 1990 which is 2014.

$P=45;000(1+r)^n$

$90;000=45;000(1+r)^{10}$

$(1+r)^{10}=dfrac{90;000}{45;000}$

$(1+r)^{10}=2$

$[(1+r)^{10}]^{frac{1}{10}}=2^{frac{1}{10}}$

$1+r=2^{frac{1}{10}}$

$r=2^{frac{1}{10}}-1$

$$
r=0.0718=7.2%
$$

b) $74;378$

c) during 2014

d) $7.2%$