I've seen a similar question asked before, but I still can't figure this out. I got $b=\frac{25}{3}$.
f(x) has an inverse if f is injective i.e. if $\frac{df}{dx} \geq 0$ for all x.
$$3x^2+10x+b \geq 0$$ $$\frac{-10^2 \pm \sqrt{10^2-4(3)(b)}}{(2)(3)} \geq 0$$ $$\sqrt{10^2-4(3)(b)} \le 0$$ $${100-12b} \le 0$$ $$b \le \frac{100}{12}$$ $$b = \frac{25}{3}$$
$f^{-1}(1) = a \Leftrightarrow (1,a)$ lies on $f^{-1}(x) \Leftrightarrow (a,1)$ lies on $f(x)$.
$$\frac{d}{dx}f^{-1}(1) = \frac{1}{f'(f^{-1}(1))}=\frac{1}{f'(a)}=\frac{1}{3a^2+10a+\frac{25}{3}}$$
But this is not an accepted answer because it contains "a", and I get the error: "Forbidden variable or constant: a." So I tried finding a value for a by setting $\frac{d}{dx}(f(a))=1$ but I end up getting two roots that are kind of ugly, and I get the wrong answer when i substitute these in for a in the expression $f^{-1}. Thanks for any answers in advance and sorry for the messy equations I could not get tabular to work (I'm new to latex and it's my first time asking a question here so any feedback appreciated).
Hint:
For what value of $x$ is $f(x)=x^3+5x^2+\frac{25}{3}x+1=1$ ?