$f(-1)=(-1)^3-3(-1)^2-10(-1)+24= -1-3(1)+10+24=-1-3+10+24=30$

$f(4)=(4)^3-3(4)^2-10(4)+24=64-3(16)-40+24=64-48-40+24=0$

#### (b)

$f(-1)=dfrac{4(-1)}{1-(-1)}=dfrac{-4}{1+1}=dfrac{-4}{2}=-2$

$f(4)=dfrac{4(4)}{1-(4)}=dfrac{16}{1-4}=dfrac{16}{-3}=-5dfrac{1}{3}$

#### (c)

$f(-1)=3 log_{10}(-1)$

Since you cannot take the log of a negative number, the expression is undefined.

$f(4)= 3 log_{10}(4)=3(0.6021)=1.81$

#### (d)

$f(-1)= -5(0.5^{(-1-1)})=-5(0.5^{-2})=-5(4)=-20$

$f(4)=-5(0.5^{4-1})=-5(0.5^3)=-5(0.125)=-0.625$

The range is the $y$-values. From the graph, the range is $left{yinBbb{R}| yne2 right}$.

There is no minimum or maximum value.
The function is never increasing.
The function is decreasing from $(-infty, 1)$ and $(1, infty)$.

The function approaches $-infty$ as $x$ approaches $1$ from the left and $infty$ as $x$ approaches $1$ from the right.
The vertical asymptote is $x=1$.
The horizontal asymptote is $y=2$.

$textbf{The vertical stretch}$ turns the function into $y=|2x|$.

$textbf{The translation}$ $3$ units right turns the function into $y=|2x-3|$.

#### (b)

$textbf{The reflection}$ in the $x$-axis turns the function into $y=-cos{x}$.

$textbf{The horizontal compression}$ by a factor of $dfrac{1}{2}$ turns the function into $y=-cos{2x}$.

#### (c)

$textbf{The reflection}$ in the $y$-axis turns the function into $y=log_{3}(-x)$.

$textbf{The translation}$ $4$ units left makes the function $y=log{3}(-x-4)-1$.

#### (d)

$textbf{The vertical stretch}$ of $4$ turns the function into $y=dfrac{4}{x}$.

$textbf{The reflection}$ in the $x$-axis turns the function into $y=-dfrac{4}{x}$.

$textbf{The vertical translation}$ $5$ units down turns the function into $y=-dfrac{4}{x}-5$.

$2x^3-7x^2-5x+4=0$

By using $textbf{synthetic division}$, we get that one solution is $x=-1$.After that, we have next:

$2x^2-9x+4=0$

$(2x-1)(x-4)=0$

$2x-1=0$ or $x-4=0$

$2x=1$ or $x=4$

$x=dfrac{1}{2}$ or $x=4$

So, $textbf{the solutions are}$ $x=-1$, $x=dfrac{1}{2}$ or $x=4$.
#### (b)

$dfrac{2x+3}{x+3}+dfrac{1}{2}=dfrac{x+1}{x-1}$ $/2(x+3)(x-1)$

$2(2x+3)(x-1)+(x+3)(x-1)=2(x+1)(x+3)$

$2(2x^2-2x+3x-3)+x^2-x+3x-3=2(x^2+3x+x+3)$

$4x^2+2x-6+x^2+2x-3=2x^2+8x+6$

$3x^2-4x-15=0$

$(3x+5)(x-3)=0$

$x=-dfrac{5}{3}$ or $x=3$

So, $textbf{the solutions are}$ $x=-dfrac{5}{3}$, $x=3$.

$log{x}+log(x-3)=1$

$log(3(x-3))=1$

$10^1=x(x-3)$

$x^2-3x-10=0$

$(x-5)(x+2)=0$

$x=5$ or $x=-2$

Cannot take the $log$ of a negative number, so $x=5$.
#### (d)

$10^{-4x}-22=978$

$10^{-4x}=1000$

$-4x=3$

$x=-dfrac{3}{4}$

So, $textbf{the solution is}$ $x=-dfrac{3}{4}$.

$5^{x+3}-5^x=0.992$

$5^3cdot5^x-5^x=0.992$

$5^x(5^3-1)=0.992$

$5^x(125-1)=0.992$

$124cdot5^x=0.992$

$5^x=0.008$

$5^x=5^-3$

$x=-3$

So, $textbf{the solution is}$ $x=-3$.

$2cos^2{x}=sin{x}+1$

$2(sin^2{x}-1)=sin{x}+1$

$2sin^2{x}-sin{x}-3=0$

$(2sin{x}-3)(sin{x}+1)=0$

$2sin{x}-3=0$ or $sin{x}+1=0$

$2sin{x}-3=0$ or $sin{x}+1=0$

$sin{x}=dfrac{3}{2}$ or $sin{x}=-1$

Since $sin{x}$ cannot be greater than $1$, the first equation does not give a solution.

$sin{x}=-1$

$x=270^circ$

So, $textbf{the solution is}$ $x=270^circ$.

(d) $x=-dfrac{3}{4}$; (e) $x=-3$; (f) $x=270^circ$.

$x^3-x^2-14x+24<0$

Find the critical points by solving $x^3-x^2-14x+24=0$.By using $textbf{synthetic division}$, we get that $2$ is a critical value.

$x^2+x-12=0$

$(x+4)(x-3)=0$

$x+4=0$ or $x-3=0$

$x=-4$ or $x=3$

$textbf{The critical values are}$ $-4$, $2$ and $3$.Test points that are in the intervals created by the critical values:

$(x-2)(x-3)(x+4)$

Test $-5$: $(-)(-)(-)=(-)0$

Test $2.5$: $(+)(-)(+)=(-)0$

So, $textbf{the solution is}$ $(-infty,-4)$ $cup$ $(2,3)$.

$dfrac{(2x-3)(x-4)}{x+2}leq0$

Find $textbf{the critical values}$:

$2x-3=0$

$2x=3$

$x=dfrac{3}{2}$

$x-4=0$

$x=4$

$x+2=0$

$x=-2$

$textbf{The critical values are}$ $dfrac{3}{2}$, $4$ and $-2$.

Test $-3: (-)(-)div(-)=(-)0$

Test $2: (+)(-)div(+)=(-)0$

$textbf{The solution is}$ $(-2,dfrac{3}{2})$ $cup$ $[4,infty)$.

$f(x)=2sin(x-pi)$

$f(-x)=2sin(-x-pi)=-2sin(x-pi)$

So,it is odd.

#### (b)

$f(x)=dfrac{3}{4-x}$

$f(-x)=dfrac{3}{4-(-x)}=dfrac{3}{4+x}$

So, it is neither.

#### (c)

$f(x)=4x^4-3x^2$

$f(-x)=4(-x)^4-3(-x)^2=4x^4-3x^2$

So, it is even.

#### (d)

$f(x)=2^{3x-1}$

$f(-x)=2^{3(-x)-1}=2^{-3x-1}$

So, it is neither.