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
Mar 5, 2017

Answer:

C) #28/5#

Explanation:

Let's first imagine what the region would look like. Inverse functions are reflections of themselves over the line #x=y#, so the lines would resemble the following (ignore any numbers, this is just a speculation):

desmos.com

We see that the top line will be given by #L_1(x)=f(x)=9/25x+b#.

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 #L_2(x)=25/9(x-b)#.

The area of the region is given by two distinct parts: the first is the trapezoid that lies under #L_1#. This extends from #x=0# to the #x# intercept of #L_2#.

The second part is a triangle that extends from the #x# intercept of #L_2# to the intersection point of #L_1# and #L_2#. Its height is the value of #L_1# at the #x# intercept of #L_2#.

Let's find these important points first:

#mathbf(L_2)# #mathbfx#-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 #b#, which is the #y# intercept of #L_1#. The other base is #L_1(b)=34/25b#. The height is #b# as well, so the trapezoid's area is

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

Triangle Area

The base is again #L_1(b)=34/25b#. The height (horizontal) is #25/16b-b=9/16b#. The area of the triangle is then

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

The total region #R# has area #49#.

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

#25/16b^2=49#

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