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lbvjy [14]
2 years ago
11

For the following exercises, determine whether the functions are even, odd, or neither.

Mathematics
1 answer:
stealth61 [152]2 years ago
6 0

Answer:

The solutions of the equation |2 x-3|=17

are x=10

and x=-7

Step-by-step explanation:

Given equations is |2 x-3|=17.

It is required to find out the values of x satisfying the given equation.

To find it out, use the fact that the solution of this type of equation is 2x-3=17.

Or 2x-3=-17 and simplify the equations using simple mathematical operations.

Step 1 of 2

Solve the equation 2x-3=17,

Add 3 on both sides of the equation 2x-3=17,

2x-3+3=17+3

2x=20

Divide by 2 on both sides of the equation 2x=20,

$$\begin{aligned}&\frac{2 x}{2}=\frac{20}{2} \\&x=10\end{aligned}$$

Step 2 of 2

Solve the equation 2x-3=-17,

Add 3 on both sides of the equation 2x-3=-17,

$2 x-3+3=-17+3$\\ $2 x=-14$

Divide by 2 on both sides of the equation 2x=-14,

$$\begin{aligned}&\frac{2 x}{2}=\frac{-14}{2} \\&x=-7\end{aligned}$$

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8_murik_8 [283]
\bf f(x)=log\left( \cfrac{x}{8} \right)\\\\&#10;-----------------------------\\\\&#10;\textit{x-intercept, setting f(x)=0}&#10;\\\\&#10;0=log\left( \cfrac{x}{8} \right)\implies 0=log(x)-log(8)\implies log(8)=log(x)&#10;\\\\&#10;8=x\\\\&#10;-----------------------------

\bf \textit{y-intercept, is setting x=0}\\&#10;\textit{wait just a second!, a logarithm never gives 0}&#10;\\\\&#10;log_{{  a}}{{  b}}=y \iff {{  a}}^y={{  b}}\qquad\qquad &#10;%  exponential notation 2nd form&#10;{{  a}}^y={{  b}}\iff log_{{  a}}{{  b}}=y &#10;\\\\&#10;\textit{now, what exponent for "a" can give  you a zero? none}\\&#10;\textit{so, there's no y-intercept, because "x" is never 0 in }\frac{x}{8}\\&#10;\textit{that will make the fraction to 0, and a}\\&#10;\textit{logarithm will never give that, 0 or a negative}\\\\&#10;

\bf -----------------------------\\\\&#10;domain&#10;\\\\&#10;\textit{since whatever value "x" is, cannot make the fraction}\\&#10;\textit{negative or become 0, , then the domain is }x\ \textgreater \ 0\\\\&#10;-----------------------------\\\\&#10;range&#10;\\\\&#10;\textit{those values for "x", will spit out, pretty much}\\&#10;\textit{any "y", including negative exponents, thus}\\&#10;\textit{range is }(-\infty,+\infty)
 p, li { white-space: pre-wrap; }

----------------------------------------------------------------------------------------------




now on 2)

\bf f(x)=\cfrac{3}{x^4}   if the denominator has a higher degree than the numerator, the horizontal asymptote is y = 0, or the x-axis,

in this case, the numerator has a degree of 0, the denominator has 4, thus y = 0


vertical asymptotes occur when the denominator is 0, that is, when the fraction becomes undefined, and for this one, that occurs at  x^4=0\implies x=0  or the y-axis

----------------------------------------------------------------------------------------------


now on 3)

\bf f(x)=\cfrac{1}{x}


now, let's see some transformations templates

\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\&#10;&#10;\begin{array}{rllll}&#10;% left side templates&#10;f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\mathbb{R}^{{{  B}}x+{{  C}}}+{{  D}}&#10;\end{array}


\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\&#10;\bullet \textit{ vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}&#10;\end{array}


now, let's take a peek at g(x)

\bf \begin{array}{lcllll}&#10;g(x)=&-&\cfrac{1}{x}&+3\\&#10;&\uparrow &&\uparrow \\&#10;&\textit{upside down}&&&#10;\begin{array}{llll}&#10;\textit{vertical shift up}\\&#10;\textit{by 3 units}&#10;\end{array}&#10;\end{array}


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3 years ago
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