One to One

We note that $f(0) = 1$ . Since $f$ is one-to-one, this implies that $f^{-1}(1) = 0$ . Let $y=f(x)$ , then $(f^{-1}(x))' = \dfrac {dx}{dy} = \dfrac 1{\frac {dy}{dx}} = \dfrac 1{21x^6+4x+4}$ . $\implies (f^{-1}(1))' =\dfrac {dx}{dy} \bigg|_{x=0} = \boxed{\dfrac 14}$ .

Since $f$ is one to one, we know the following property holds for $f$ and its inverse: $f(f^{-1}(x)) = x$ and differentiating both sides gives $f'(f^{-1}(x))(f^{-1}(x))' = 1$ $(f^{-1}(x))' = \frac{1}{ f'(f^{-1}(x))}$ and all that remains is to compute $f^{-1}(1)$ . To do this we notice that $f(0) = 1$ so it must be that $0 = f^{-1}(1)$ and since $f'(0) = 4$ we have $(f^{-1}(1))' = \boxed{\frac{1}{4}}$