# precalc!!!!!

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The point (5,-2) lies on the graph of y=f(x), and assume f^-1(x) exists. Find the corresponding point on eAch graph.

A). Y=f^-1(x)

I get (-5,2)

B). Y=-f^-1(-x)+4
I get (5,2)

C). Y=f^-1(x+2)+3

I get (-7,5)

Did I get these right?

• precalc!!!!! -

say y = x - 7
(5,-2) lies on that graph
for f^-1
x = y - 7
y = x + 7
if x = 5 then y = 12
so I get
(5,12)

• dumb mistake -

say y = x - 7
(5,-2) lies on that graph
for f^-1
x = y + 7
y = x - 7
if x = 5 then y = -2
so I get
(5,-2)

• precalc!!!!! -

B. My f^-1(x) = x-7
Y=-f^-1(-x)+4
Y = (-5-7) + 4
Y = -12+ 4 = -8
(5,-8)

C.Y=f^-1(x+2)+3
x+2 = 7
f^-1(7) = 7-7 = 0
Y = 0 + 3 = 3

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