# Show that g(x)=1/(x-1)+1/(x-2) given g(x)=f'(x)/f(x), f(x)=(x-1)(x-2)?

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You need to evaluate the original form of the function `g(x)` , hence, since its numerator is given by the first order derivative of the function `f(x)` , you need to differentiate the function `f(x)` with respect to `x` , using product rule, such that:

`f'(x) = ((x-1)(x-2))' => f'(x) = (x-1)'(x-2) + (x-1)(x-2)'`

`f'(x) = x - 2 + x - 1 => f'(x) = 2x - 3`

`g(x) = (f'(x))/(f(x)) => g(x) = (2x - 3)/((x-1)(x-2))`

You need to test if the function `g(x) =(2x - 3)/((x-1)(x-2))` can be converted to the form `g(x) = 1/(x-1) + 1/(x-2)` , hence, you need to use partial fraction expansion, such that:

`(2x - 3)/((x-1)(x-2)) = a/(x - 1) + b/(x - 2)`

Bringing the terms to a common denominator yields:

`2x - 3 = a(x - 2) + b(x - 1)`

`2x - 3 = ax - 2a + bx - b`

`2x - 3 = x(a + b) - 2a - b`

Equating the coefficents of like powers yields:

`{(a + b = 2),(-2a - b = -3):} => `

Adding the equations yields:

`-a = -1 => a = 1 => 1 + b = 2 => b = 1`

Replacing 1 for a and b yields:

`(2x - 3)/((x-1)(x-2)) = 1/(x - 1) + 1/(x - 2)`

**Hence, the last line proves that the equation of the function g(x) can be converted into `g(x) = 1/(x - 1) + 1/(x - 2)` .**