Question

If $f\left(x\right)=2x^3+7x-5$ then $f^{-1}\left(4\right)$ is

• A. Equal to $1$
• B. Equal to $2$
• C. Equal to $1/3$
• D. Non existent

Answer 1

Answer: (A)

Equal to $1$

$f\left(x\right)=2x^3+7x-5$

$x=2\left({f^{-1}\left(x\right)}^3\right)+7f^{-1}\left(x\right)-5$

$f^{-1}\left(4\right)=?$

$4=2\left({f^{-1}\left(4\right)}^3\right)+7f^{-1}\left(4\right)-5$

$2\left({f^{-1}\left(4\right)}^3\right)+7f^{-1}\left(4\right)-9=0$

Let $f^{-1}\left(4\right)=t$

$2t^3+7t-9=0$

$2t^3-2+7t-7=0$

$2(t^3-1)+7(t-1)=0$

$2(t-1)(t^2+t+1)+7t-1=0$

$(t-1)(2t^2+2t+2+7)=0$

$t=12t^2+2t+9=0$

So, $f^{-1}\left(4\right)=1$

Answer A