# 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

Mar 5, 2017

C) $\frac{28}{5}$

#### 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):

We see that the top line will be given by ${L}_{1} \left(x\right) = f \left(x\right) = \frac{9}{25} x + b$.

The inverse of this is given by:

$y = \frac{9}{25} x + b \text{ "=>" } x = \frac{9}{25} y + b$

$\textcolor{w h i t e}{y = \frac{9}{25} x + b} \text{ "=>" } y = \frac{25}{9} \left(x - b\right) = {f}^{-} 1 \left(x\right)$

Since this is the lower line we say that ${L}_{2} \left(x\right) = \frac{25}{9} \left(x - b\right)$.

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:

$m a t h b f \left({L}_{2}\right)$ $m a t h b f x$-intercept

${L}_{2} \left(x\right) = \frac{9}{25} \left(x - b\right) = 0 \text{ "=>" } x = b$

Intersection Point

${L}_{1} \left(x\right) = {L}_{2} \left(x\right) \text{ "=>" } \frac{9}{25} x + b = \frac{25}{9} \left(x - b\right)$

$\textcolor{w h i t e}{{L}_{1} \left(x\right) = {L}_{2} \left(x\right)} \text{ "=>" } \frac{81}{625} x + \frac{9}{25} b = x - b$

$\textcolor{w h i t e}{{L}_{1} \left(x\right) = {L}_{2} \left(x\right)} \text{ "=>" } \frac{34}{25} b = \frac{544}{625} x$

$\textcolor{w h i t e}{{L}_{1} \left(x\right) = {L}_{2} \left(x\right)} \text{ "=>" } x = \frac{25}{16} b$

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 ${L}_{1} \left(b\right) = \frac{34}{25} b$. The height is $b$ as well, so the trapezoid's area is

${A}_{1} = \frac{1}{2} b \left(b + \frac{34}{25} b\right) = \frac{59}{50} {b}^{2}$

Triangle Area

The base is again ${L}_{1} \left(b\right) = \frac{34}{25} b$. The height (horizontal) is $\frac{25}{16} b - b = \frac{9}{16} b$. The area of the triangle is then

${A}_{2} = \frac{1}{2} \left(\frac{34}{25} b\right) \left(\frac{9}{16} b\right) = \frac{153}{400} {b}^{2}$

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

$R = {A}_{1} + {A}_{2} = \frac{59}{50} {b}^{2} + \frac{153}{400} {b}^{2} = 49$

$\frac{25}{16} {b}^{2} = 49$

$b = \sqrt{\frac{49 \left(16\right)}{25}} = \frac{28}{5}$