pa18
contestada

Part 1.] Determine the equation of the inverse of [tex]y= \frac{1}{4} x^{3} -2[/tex]
A.] [tex]y= \sqrt[3]{4x+2} [/tex]
B.] [tex]y= \sqrt[3]{4x-2} [/tex]
C.] [tex]y= \sqrt[3]{4x+8} [/tex]
D.] [tex]y= \sqrt[3]{4x-8} [/tex]
E.] [tex]y= \sqrt[3]{ \frac{1}{4}x-2} [/tex]

Part 2.] Determine the equation of the inverse of [tex]y=e^{x+3} -4[/tex].
A.] [tex]y=ln(x+4)-3[/tex]
B.] [tex]y=ln x+1[/tex]
C.] [tex]y= \frac{1}{3}( \frac{x}{e}+4) [/tex]
D.] [tex]y= \frac{1}{4}lnx-3[/tex]

Respuesta :

y = 1/4 x^3 + 2
1/4 x^3 = y - 2
x^3 =  4(y - 2)

x = cube root (4y - 8)

inverse y = cube root (4x - 8)

Its  D.