Question

Elasticity of a function `("E"y)/("E"x)` is given by `("E"y)/("E"x) = (-7x)/((1 - 2x)(2 + 3x))`. Find the function when x = 2, y = `3/8`

Answer 1

`("E"y)/("E"x) = (-7x)/((1 - 2x)(2 + 3x))`

`x/y ("d"y)/("d"x) = (-7x)/((1 - 2x)(2 + 3x))`

`1/y "d"y = (-x)/(x(1 - 2x)(2 + 3x))  "d"x`

`1/y "d"y = (-7)/((1 - 2x)(2 + 3x))  "d"x`

`1/y "d"y = 7/((2x - 1)(3x + 2))  "d"x`  .......(1)

Let `7/((2x + 1)(3x + 2)) = "A"/((2x - 1)) + "B"/((3x + 2))`

`7/((2x - 1)(3x + 2)) = ("A"(3x + 2) + "B"(2x - 1))/((2x - 1)(3x + 2))`

7 = A(3x + 2) + B(2x – 1)

Put x = `1/2`

7 = `"A"(3(1/2) + 2) + "B"(2(1/2) - 1)`

7 = `"A"(3/2 + 2) + "B"(1 - 1)`

7 = `"A"((3 + 4)/2) + "B"(0)`

7 = `"A"(7/2)`

⇒ A = 2

Put x = 0

7 = A(3(0) + 2) + B(2(0) – 1)

7 = A(2) + B(– 1)

7 = (2)(2) – B

B = 4 – 7

B = – 3

⇒ `1/y  "d"y = 7/((2x - 1)(3x + 2))  "d"x`

`1/y "d"y = [2/((2x - 1)) - 3/((3x + 2))]  "d"x`

Integrating on both sides

`int 1/y "d"y = int 2/((2x - 1))  "d"x - int 3/((3x + 2))  "d"x`

`log |y| = log |2x - 1| - log |3x + 2| + log "k"`

log [y] = `log (("k"(2x - 1))/((3x + 2)))`

⇒ y = `("k"(2x - 1))/((3x + 2))`

⇒ (2)

When x = 2

y = `3/8`

`3/8 = ("k"[2(2) - 1])/[3(2) + 2]`

= `("k"[3])/8`

k = `3/8 xx 8/3`

⇒ k = 1

Equation (2)

∴ y = `((2x - 1))/((3x + 2))`