# Let #R# be the region in the first quadrant bounded by the #x# and #y# axis and the graphs of #f(x) = 9/25 x +b# and #y = f^-1 (x)#. If the area of #R# is 49, then the value of #b#, is ? A) #18/5# B) #22/5# C) #28/5# D) none

##### 1 Answer

C)

#### Explanation:

Let's first imagine what the region would look like. Inverse functions are reflections of themselves over the line

We see that the top line will be given by

The inverse of this is given by:

#y=9/25x+b" "=>" "x=9/25y+b#

#color(white)(y=9/25x+b)" "=>" "y=25/9(x-b)=f^-1(x)#

Since this is the lower line we say that

The area of the region is given by two distinct parts: the first is the trapezoid that lies under

The second part is a triangle that extends from the

Let's find these important points first:

**-intercept**

#L_2(x)=9/25(x-b)=0" "=>" "x=b#

**Intersection Point**

#L_1(x)=L_2(x)" "=>" "9/25x+b=25/9(x-b)#

#color(white)(L_1(x)=L_2(x))" "=>" "81/625x+9/25b=x-b#

#color(white)(L_1(x)=L_2(x))" "=>" "34/25b=544/625x#

#color(white)(L_1(x)=L_2(x))" "=>" "x=25/16b#

We can now, without any integration necessary (since we're dealing with lines) find these areas.

**Trapezoid Area**

One base is

#A_1=1/2b(b+34/25b)=59/50b^2#

**Triangle Area**

The base is again

#A_2=1/2(34/25b)(9/16b)=153/400b^2#

The total region

#R=A_1+A_2=59/50b^2+153/400b^2=49#

#25/16b^2=49#

#b=sqrt((49(16))/25)=28/5#