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 Pre-Calculus and Calculus
 inverse function question
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jst706
New Member

USA
1 Posts

Posted - 09/25/2007 :  20:09:07  Show Profile
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^-4-x^-2

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

h'(2)=1/(-3(16)-4)

Thanks for any help I can get
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skeeter
Advanced Member

USA
5634 Posts

Posted - 09/25/2007 :  22:17:32  Show Profile
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
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Mrspi
Advanced Member

USA
998 Posts

Posted - 09/25/2007 :  23:06:30  Show Profile
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^-4-x^-2

= 1/ (-3x^4-x^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) = x3 + x

then the inverse of f(x) is NOT 1/(x3 + x)

let y = f(x)

y = x3 + x

Interchange x and y.....

x = y3 + y

Now...solve for y.....

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