Find a formula for the inverse of the following function, if possible.

Question

KyledaGreat · Answer

$$V(x) = 2x ^{\frac{ 1 }{ 5}} + 3$$

$$V^{-1}(x) = $$

Florisalreadytaken · Answer

$ V(x)=2x^{\frac{1}{5}}+3  $

we can use $y$ for $V(x)  $:

$y=2x^{\frac{1}{5}}+3$

$ y-3=2x^{\frac{1}{5}}  $
$$ y=2x^{\frac{1}{5}}+3 $$
$$  \left(x^{\frac{1}{5}}\right)^5=\left(\frac{y-3}{2}\right)^5 $$
$$ x^{\frac{1}{5}\cdot \frac{5}{1}}=\frac{\left(y-3\right)^5}{32}  $$
$$ x=\frac{\left(y-3\right)^5}{32}  $$

if you do not have any info on how to solve the above equation, this is how you do it, but if your teacher hasn't taught you about it  yet, just use the above equation as the answer

here we have to use the binomial theorem    $ \ \  \left(a\pm b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i \ \  $    to expand the  numerator which is in the form of $  (a \pm b)^n   $

so for $  a=y,\:\:b=-3    $ we get    $ \sum _{i=0}^5\binom{5}{i}y^{\left(5-i\right)}\left(-3\right)^i $ where $ \binom{n}{i}=\frac{n!}{i!\left(n-i\right)!} $

lets start with the first one, then second then third and so on:

$$ \:i=0\quad :\quad \frac{5!}{0!\left(5-0\right)!}y^5\left(-3\right)^0  \ \ \ = \ \ \frac{5!}{1\cdot \:5!}\cdot \:y^5\cdot 1 \ \ = \ \ \frac{5!}{5!} \:y^5 \ \ = \ \ 1\cdot \:y^5 = \boxed{y^5} $$
$$ \:i=1\quad :\quad \frac{5!}{1!\left(5-1\right)!}y^4\left(-3\right)^1 \ \ = \ \  y^4 \frac{5!}{1!\cdot \:4!}y^4 \ \ \overset{\frac{5!}{4!}=5}{\Rightarrow} \ \ -\frac{5y^4\cdot \:3}{1!} \ \ =\boxed{ -15y^4} $$

did the same for other ones with `www.wolframalpha.com/` and got the results, cuz it would take too much time -- you get the idea of how to do it already:

$$  i=2\quad :\quad \frac{5!}{2!\left(5-2\right)!}y^3\left(-3\right)^2 \ \  = \ \  \boxed{90y^3} $$
$$  i=3\quad :\quad \frac{5!}{3!\left(5-3\right)!}y^2\left(-3\right)^3  \ \ = \ \ \boxed{-270y^2} $$
$$ i=4\quad :\quad \frac{5!}{4!\left(5-4\right)!}y^1\left(-3\right)^4 \ \ = \ \ \boxed{405y} $$
$$  i=5\quad :\quad \frac{5!}{5!\left(5-5\right)!}y^0\left(-3\right)^5 \ \ = \ \ \boxed{-243} $$

adding it alltogether we get $ (y-3)^5=y^5-15y^4+90y^3-270y^2+405y-243 $

lets plug that into the above equation:

$$ x^1=\frac{y^5-15y^4+90y^3-270y^2+405y-243}{2^5} $$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Updownarrow   $$
$$x=\frac{y^5}{32}-\frac{15y^4}{32}+\frac{90y^3}{32}-\frac{270y^2}{32}+\frac{405y}{32}-\frac{243}{32}  $$

did this wecause we got some oppertunitis to cancel like $ \frac{90y^3}{32} $ and $ \frac{270y^2}{32} $:

$$ x=\frac{y^5}{32}-\frac{15y^4}{32}+\frac{\cancel{90}y^3}{\cancel{32}}-\frac{\cancel{270}y^2}{\cancel{32}}+\frac{405y}{32}-\frac{243}{32} $$

$$  x=\frac{y^5}{32}-\frac{15y^4}{32}+\frac{45y^3}{16}-\frac{135y^2}{16}+\frac{405y}{32}-\frac{243}{32} $$

That done, we can  say that the final answer is:

$$  \boxed{\:V^{-1}(x)=\frac{y^5-15y^4+405y-243}{32}+\frac{45y^3-135y^2}{16} }$$

snowflake0531 · Answer

replace x with y and y with x to find the inverse of a function
$$y=2x^{\frac{1}{5} }+3 $$
 |
 v
$$x=2y^{\frac{1}{5}}+3$$
then solve for y

$$ x-3 = 2y^{\frac{1}{5}}\\frac{x-3}{2}=y^{\frac{1}{5}}\y=(\frac{x-3}{2})^5$$

snowflake0531 · Answer

so the inverse $$ V^{-1}(x)=(\frac{x-3}{2})^5$$