Posted by **L.Bianchessi** on Saturday, March 10, 2012 at 9:14pm.

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!!!!! -
**Damon**, Sunday, March 11, 2012 at 6:45am
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 -
**Damon**, Sunday, March 11, 2012 at 6:48am
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!!!!! -
**Damon**, Sunday, March 11, 2012 at 7:00am
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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