Strange problem on inverse function =S

TWiX · #1 $Old$ June 3rd, 2009, 02:38 AM

hi

am solving past exams to be prepared for my exam next week
i stopped on this problem :

let f(x)=x^4 + x^3 + 1 , where x in [0,2] let g(x) = f^-1 (x)

if h(x) = f(2 g(x)) , find h`(3)??

its easy problem, i know all what i know is how to find h`(x) ??
then it will be easy

help plz !!

HallsofIvy · #2 $Old$ June 3rd, 2009, 05:22 AM

Quote:

Originally Posted by TWiX $View Post$

hi

am solving past exams to be prepared for my exam next week
i stopped on this problem :

let f(x)=x^4 + x^3 + 1 , where x in [0,2] let g(x) = f^-1 (x)

if h(x) = f(2 g(x)) , find h`(3)??

its easy problem, i know all what i know is how to find h`(x) ??
then it will be easy

help plz !!

I presume that h`(x) is the derivative. (I would have used h'(x).)

Use the chain rule:
h'(x)= 2g'(x) f '(2g(x)) and g'(x)= 1/f '(x)

It is easy to calculate that $f'(x)= 4x^3+ 3x^2$

The only "hard" part, perhaps, is finding g(3). To do that you have to solve $f(x)= x^4+ ^3+ 1= 3$ so $x^3+ x^2= 2$ . Fortunately that has an easy soution in [0, 1]!

TWiX · #3 $Old$ June 3rd, 2009, 10:33 AM

it seems we will have to use the composite rule to differentiate it !?!

HallsofIvy · #4 $Old$ June 3rd, 2009, 10:43 AM

Quote:

Originally Posted by TWiX $View Post$

it seems we will have to use the composite rule to differentiate it !?!

Yes, of course! It is a composite function.

TWiX · #5 $Old$ June 3rd, 2009, 11:10 AM

What is the formula of the composite rule ??

Amer · #6 $Old$ June 3rd, 2009, 11:27 AM

$h(x) = f(2 g(x))$

$h'(x) =\left( \frac{d}{dx}(2g(x))\right)f'(2g(x))$

$h'(x) = 2g'(x)f'(2g(x))$

TWiX · #7 $Old$ June 3rd, 2009, 01:24 PM

Thank you ;]]

TWiX · #8 $Old$ June 4th, 2009, 09:41 AM

What is the domain of the composites Function ??

suppose the doman of f = A
domain of g = B

what is the doman of f( g(x) ) , g( f(x) ) ?!

Amer · #9 $Old$ June 4th, 2009, 10:45 AM

$let ...f:A\rightarrow B...G:C\rightarrow D$

so

$let..h=f(g)...h:A\rightarrow D$

so

f(g(x)) domain f but range of g
g(f(x)) domain g but range of f

skeeter · #10 $Old$ June 4th, 2009, 12:05 PM

if $g(x) = f^{-1}(x)$ , then ...

$f[g(x)] = g[f(x)] = x$

$f'[g(x)] \cdot g'(x) = g'[f(x)] \cdot f'(x) = 1$

so ...

$f'[g(x)] = \frac{1}{g'(x)}$

and

$g'[f(x)] = \frac{1}{f'(x)}$

$f(x) = x^4 + x^3 + 1$ and $f'(x) = 4x^3 + 3x^2$

note that $f(1) = 3$ , so $g(3) = 1$

$g'(3) = g'[f(1)] = \frac{1}{f'(1)} = \frac{1}{7}$

$h'(x) = f'[2g(x)] \cdot 2g'(x)$

$h'(3) = f'[2g(3)] \cdot 2g'(3) = f'[2(1)] \cdot 2\left(\frac{1}{7}\right) = \frac{88}{7}$