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jst706
New Member

USA
1 Posts

 Posted - 09/25/2007 :  20:09:07 the question is....Let f(x)=x^3+x. If h is the inverse function of f, then h'(2)=the answer is 1/4, but I dont understand how to do it..this problem is from the collegeboard clep guide and it doesn't explain any problems. All I know is I went horribly wrong somewhere....this is what I did.1/(x^3+x)...I start with that, then I take the derivative of thatx^-3+x^-1-3x^-4-x^-2= 1/ (-3x^4-x^2)h'(2)=1/(-3(16)-4)Thanks for any help I can get

skeeter

USA
5634 Posts

 Posted - 09/25/2007 :  22:17:32 if f(x) and h(x) are inverses, then f[h(x)] = x.take the derivative of the above equation w/r to x ...f'[h(x)]*h'(x) = 1so ...h'(x) = 1/f'[h(x)]therefore ...h'(2) = 1/f'[h(2)]the "trick" is to find h(2). since f(x) = x + x, we need to find x such that f(x) = 2 ...x + x = 2x + x - 2 = 0by "inspection", it's rather obvious that x = 1 is a solution ... so, f(1) = 2, and since h is the inverse of f, then h(2) = 1.now go back to the derivative equation h'(2) = 1/f'[h(2)] ...h'(2) = 1/f'(1) ... f'(x) = 3x + 1 -> f'(1) = 4finally ...h'(2) = 1/4

Mrspi

USA
998 Posts

 Posted - 09/25/2007 :  23:06:30 quote:Originally posted by jst706the question is....Let f(x)=x^3+x. If h is the inverse function of f, then h'(2)=the answer is 1/4, but I dont understand how to do it..this problem is from the collegeboard clep guide and it doesn't explain any problems. All I know is I went horribly wrong somewhere....this is what I did.1/(x^3+x)...I start with that, then I take the derivative of thatx^-3+x^-1-3x^-4-x^-2= 1/ (-3x^4-x^2)h'(2)=1/(-3(16)-4)Thanks for any help I can getHere's something to think about for "starters".....if f(x) = x3 + xthen the inverse of f(x) is NOT 1/(x3 + x)let y = f(x)y = x3 + xInterchange x and y.....x = y3 + yNow...solve for y.....
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