Mathematics, 09.12.2021 23:20 NikkiZoeller

How do I find the inverse?

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Answer from: yasarhan2

g(x) as given has no inverse because there are instances of two x values giving the same value of g(x). For instance,

x = -1 ⇒ g(-1) = 4 (-1 + 3)² - 8 = 8

x = -5 ⇒ g(-5) = 4 (-5 + 3)² - 8 = 8

Only a one-to-one function can have an inverse. g(x) is not one-to-one.

However, if we restrict the domain of g(x), we can find an inverse over that domain. Let $g^{-1}(x)$ be the inverse of g(x). Then by definition of inverse function,

$g\left(g^{-1}(x)\right) = 4 \left(g^{-1}(x) + 3\right)^2 - 8 = x$

Solve for the inverse:

$4 \left(g^{-1}(x) + 3\right)^2 - 8 = x$

$4 \left(g^{-1}(x) + 3\right)^2 = x + 8$

$\left(g^{-1}(x) + 3\right)^2 = \dfrac{x + 8}4$

$\sqrt{\left(g^{-1}(x) + 3\right)^2} = \sqrt{\dfrac{x + 8}4}$

$\left| g^{-1}(x) + 3 \right| = \dfrac{\sqrt{x+8}}2$

Recall the definition of absolute value:

$|x| = \begin{cases}x & \text{if }x\ge0\\-x&\text{if }x$

This means there are two possible solutions for the inverse of g(x) :

• if $g^{-1}(x) + 3 \ge 0$ , then

$g^{-1}(x) + 3 = \dfrac{\sqrt{x+8}}2 \implies g^{-1}(x) = -3+\dfrac{\sqrt{x+8}}2$

• otherwise, if $g^{-1}(x)+3$ , then

$-\left(g^{-1}(x) + 3\right) = \dfrac{\sqrt{x+8}}2 \implies g^{-1}(x) = -3-\dfrac{\sqrt{x+8}}2$

Which we choose as the inverse depends on how we restrict the domain of g(x). For example:

Remember that the inverse must satisfy

$g\left(g^{-1}(x)\right) = x$

In the first case above, $g^{-1}(x) + 3 \ge 0$ , or $g^{-1}(x) \ge -3$ . This suggests that we could restrict the domain of g(x) to be $x \ge -3$ .

Then as long as $x \ge -3$ , the inverse is

$g^{-1}(x) = -3+\dfrac{\sqrt{x+8}}2$

Answer from: Quest

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

ver más en -

step-by-step explanation:

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