Author 
Topic 

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 that
x^3+x^1 3x^4x^2
= 1/ (3x^4x^2)
h'(2)=1/(3(16)4)
Thanks for any help I can get 


skeeter
Advanced Member
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) = 1
so ...
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 = 2
x + x  2 = 0
by "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) = 4
finally ...
h'(2) = 1/4 


Mrspi
Advanced Member
USA
998 Posts 
Posted  09/25/2007 : 23:06:30

quote: Originally posted by jst706
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 that
x^3+x^1 3x^4x^2
= 1/ (3x^4x^2)
h'(2)=1/(3(16)4)
Thanks for any help I can get
Here's something to think about for "starters".....
if f(x) = x^{3} + x
then the inverse of f(x) is NOT 1/(x^{3} + x)
let y = f(x)
y = x^{3} + x
Interchange x and y.....
x = y^{3} + y
Now...solve for y.....




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