f(x) = (4x + 1)/(x - 2) with x > 2. Find a value for 'x' such that f'(x) (first derivative of f(x) with respect to x) is equal to -1.

  1. Re-write f(x)  using indicies, so that the use of the product rule becomes more clear => f(x) = (4x + 1) *(x - 2)^-1 where ^ means 'raised to the power of' and * means 'multiplied by'.

  2. use the product rule to differentiate f(x) => f'(x) = [(x - 2)^-1*d/dx(4x + 1)] + [(4x + 1)*d/dx(x - 2)^-1]. In order to work out the derivative in the second term 'd/dx(x - 2)^-1', it is a good idea to use a substitution since (x - 2)^-1 is a function of x raised to a power so it does not change linearly with x as the derivative in the first term does, so is a little more complicated to evaluate. 

  3. use z(x) = (x - 2) which turns the derivative in the second term into d/dx(z(x)^-1). If some function g(z) = z(x)^n, then dg(z)/dx = dz(x)/dx * dg(z)/dz, since g is a function of z, and z is a function of x.

applying this to the derivative from part 2, this gives d/dx(x - 2)^-1 = (-1) * (x - 2)^2 * (1)

  1. Writing out this along with the other derivative terms in the eqution from part 2 gives f'(x) = -9/(x - 2)^2.

  2. equation this to -1 and rearranging for x gives the answer. 

Answered by Zak S. Maths tutor

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