Consider the function $f(x)=x^7+x^5+3$. Show that the inverse $g(x)$ of $f(x)$ exist and calculate $F'(5)$ if $F(x)=\int_x^{-3} g(t) dt$.
I already showed that $g(x)$ exists y showing that $f(x)$ is strictly increasing (the only critical point was inflection point). However I failed finding $g(x)$ explicitly (and as the problem is formulated I think the idea is to do not find it).
What I tried was, since a function and its inverse, are $(x,f(x))$ and $(g(x),x)$, then they look as reflections over the axis $y=x$. So, notice that $f(x)$ is above the $y=x$ line for $x\geq 3$ then $g(x)$ is under the $y=x$ axis for $x\leq 3$.
So I used the symmetry to stablish that $$F(x)=\int_x^{-3} g(t) dt = -\int_{3}^{-x} f(t) dt = \int_{-x}^{3} f(t) dt = C - \left(\frac{x^8}{8} + \frac{x^6}{6}-3x \right)$$
And so $F'(x)=-x^{7}-x^{6}+3$, but in $F'(5)$ I get a negative number, and I don't get why this would make sense. Did I messed up a sign somewhere? Is my reasoning wrong?
Using the fundamental theorem of calculus, you get $F'(x)=-g(x)$ So $F'(5)=-g(5)$
So you are down to solving for x for $5=x^7+x^3+3$, which by inspection is $x=1$, thus $F'(5)=-1$.
Your problem was subbing in $f$ for $g$ in the integral before applying the FTC