Math Is Fun Forum

Neela

Guest

Functions

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I need help.

Consider the function h : x --> (x-2) / (x-1)^2

1) Find h'(x), writing the answer in the form (a-x) / (x-1)^n where a and n are constants to be determined.

2) given that h''(x)= (2x-8) / (x-1)^4 calculate the coordinates of P (point of inflexion on righthand side)

and

f(x)= 3x^4 - 4x^3 - 30x^2 -36x +112, -2 _< x _<4,5.

given that f(x) = o has 1 solution at x=4, find the other solution

The tangent to the graph of f is horizontal at x=3 and at one other value of x. find this other value.

THANK YOU SO MUCH!

Dilbert

Member

Registered: 2006-09-05

Posts: 12

Re: Functions

You need to make clear your question by the use of brackets first, etc.
The first one you use the quotient rule.

y= (x-2) / (x-1)^2

so y(x) = u(x)/v(x)

then 

dy/dx = (v(du/dx) - u(dv/dx))/ v^2

or y' = (vu' - uv')/v^2

u = (x-2), v = (x-1)(x-1) = (x^2-2x+1) (we open up brackets to get to simple orders)

so, du/dx = 1, dv/dx = 2x-2

Therefore: dy/dx = ((x^2-2x+1)*1 - (x-2)*(2x-2))/ (x^2-2x+1)^2

= (x^2-2x+1)-2x^3+2x+4/(x^2-2x+1)^2

= (-2x^3 +x^2 + 5)/(x^2 - 2x +1)^2

= (-2x^3 +x^2  +5)/x^4 - 4x^3 +2x^2  - 4x +1

I could be very wrong though.

Dross

Member

Registered: 2006-08-24

Posts: 325

Re: Functions

You can get a much nicer solution to (1) if you write h(x) as (x-2)[(x-1)^-2] and differentiate using the product rule:

So in this case, a = n = 3.

As for (2), the point of inflextion occurs where h''(x) = 0.

For the last question, since you know that x = 4 is a solution you know that the expression for f(x) has a factor of (x-4) in it. So it can be re-written as:

for some a, b, c, d. Find these and then solve

"The tangent of f at x = 3 is horizontal" is just telling you that f'(3) = 0. Given this, find f'(x), use the method I just mentioned to take out a factor of (x-3) and you should be left with a quadratic to solve to get f'(x) = 0.

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