# Thread: Strange problem on inverse function =S

1. ## Strange problem on inverse function =S

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 !!

2. Originally Posted by TWiX
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]!

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

4. Originally Posted by TWiX
it seems we will have to use the composite rule to differentiate it !?!
Yes, of course! It is a composite function.

5. What is the formula of the composite rule ??

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

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

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

7. Thank you ;]]

8. 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) ) ?!

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

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

10. 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}
$