Question: Differentiation

Megatron_X

f(x)= (x+1)^-1 so then f '(x)= -(x+1)^-2

I had to get the two asymptotes (x=-1, y=0) and I proved how there are now turning points or a point of inflextion.

The last part of the question said, the two curves are parallel at x=x1 and at x=x2. Show that x1 + x2 + 2 =0, x1[URL="javascript:;"][/URL]x2

What I did was:
x1=-x2-2

They are parallel which means that f'(x1)=f'(x2). I put the values in and in the end showed that f'(x1) = f'(x2). Does showing they are parallel by using that equation prove show that the equation is true? If not, how am I supposed to do it?

MathsManiac

Megatron_X wrote: »

f(x)= (x+1)^-1 so then f '(x)= -(x+1)^-2

I had to get the two asymptotes (x=-1, y=0) and I proved how there are now turning points or a point of inflextion.

The last part of the question said, the two curves are parallel at x=x1 and at x=x2. Show that x1 + x2 + 2 =0, x1[URL="javascript:;"][/URL]x2

What I did was:
x1=-x2-2

They are parallel which means that f'(x1)=f'(x2). I put the values in and in the end showed that f'(x1) = f'(x2). Does showing they are parallel by using that equation prove show that the equation is true? If not, how am I supposed to do it?

Strictly speaking, you showed "If A then B" when you were asked to show "If B then A". But you should be able to adapt your solution to cover that. (since all the steps in your argument are probably "reversible", more or less.

Your technique is correct other than this. So can you do it the other way: start with f'(x1) = f'(x2) and see if you can end back at {x1=x2 or x1+x2=2}.