1. inverse functions

ok i have a function f(x)= x^4+x^3+1 that i need to find the inverse of, only thing is that when i switch the x and the y to solve, it gets real messy. any help would be appreciated. thanks

2. why do you HAVE to find the inverse of that function? if it's for derivatives there is a rule for the derivative of the inverse

3. Well it depends on the domain you choose.The function as it is, is a MANY-ONE and INTO function and therefore you cannot find its inverse.You need to choose a suitable domain and corresponding codomain so that the function could be ONE-ONE and ONTO.
Even then you end up only proving existence of inverse and may not be able to get the actual inverse.

4. inverse functions

im sorry the domain is [0,2] the function is f(x) = x^4 + x^3 + 1. i need to find the inverse of this function and im having a hard time doing so. any help would be appreciated. thanks.

5. Originally Posted by artvandalay11
why do you HAVE to find the inverse of that function? if it's for derivatives there is a rule for the derivative of the inverse
i saw the formula for finding the derivative of the inverse, so are you saying i can use this, and then intergrate my answer to get the inverse function?

6. Originally Posted by slapmaxwell1
im sorry the domain is [0,2] the function is f(x) = x^4 + x^3 + 1. i need to find the inverse of this function and im having a hard time doing so. any help would be appreciated. thanks.
Please post the whole question. Usually when an inverse function is difficult to find, it turns out that it doesn't actually have to be found ....

7. Originally Posted by mr fantastic
Please post the whole question. Usually when an inverse function is difficult to find, it turns out that it doesn't actually have to be found ....
ok i see what you mean, i think i got it. i was a lil tired when i attempted this problem. i do apologize for the inconvience i may have caused. here is the original problem. let f(x)=x^4 + x^3 + 1, [0,2]

a) show that f is one to one
b) let g(x) = the inverse of f(x) and define F(x) = f(2g(x)). find an equation for the tangent line to y = F(x) at x = 3.

so to find the equation of a line i just need the point and the slope. first in getting the slope i use the chain rule to give me F'(x) = f'(2g(x)) * 2g'(x) * g'(x)? i think well i hope i executed the chain rule properly, its been a while since ive used it. but i think once i find the slope then i would plug in 3 to get the y-value and then substitute into the point slope formula?

8. Originally Posted by slapmaxwell1
first in getting the slope i use the chain rule to give me F'(x) = f'(2g(x)) * 2g'(x) * g'(x)? i think well i hope i executed the chain rule properly, its been a while since ive used it.
No

$(foh)' = f'oh \times h'$

Here $h = 2g$ then $h' = 2g'$

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

$F'(x) = f'(2g(x)) \times 2g'(x)$

Now $g=f^{-1}$ then $g' = \frac{1}{f'of^{-1}}$

$g'(3) = \frac{1}{f'of^{-1}(3)} = \frac{1}{f'(1)}$ because $f^{-1}(3) = 1$

$f'(x) = 4x^3 + 3x^2$ therefore $f'(1) = 7$ and $g'(3) = \frac17$

I think that you can finish now

9. Originally Posted by slapmaxwell1
ok i see what you mean, i think i got it. i was a lil tired when i attempted this problem. i do apologize for the inconvience i may have caused. here is the original problem.
Please post the complete problems in future