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
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.
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?
No
$\displaystyle (foh)' = f'oh \times h'$
Here $\displaystyle h = 2g$ then $\displaystyle h' = 2g'$
$\displaystyle F(x) = f(2g(x))$
$\displaystyle F'(x) = f'(2g(x)) \times 2g'(x)$
Now $\displaystyle g=f^{-1}$ then $\displaystyle g' = \frac{1}{f'of^{-1}}$
$\displaystyle g'(3) = \frac{1}{f'of^{-1}(3)} = \frac{1}{f'(1)}$ because $\displaystyle f^{-1}(3) = 1$
$\displaystyle f'(x) = 4x^3 + 3x^2$ therefore $\displaystyle f'(1) = 7$ and $\displaystyle g'(3) = \frac17$
I think that you can finish now