# Thread: [SOLVED] Stuck on a couple of problems

1. ## [SOLVED] Stuck on a couple of problems

$y^3 + y = e^x -x + 9$

a) Show that for every x, there exists only exactly one y.

b) Let the solution $y$ in (a) be $y = f(x).$ Show that this function is $C^1$ locally.

c) With the previous two answers in mind, how does it now follow that $f \in C^1(R)$

d) What is the domain and range of f.

2. Originally Posted by Spec
$y^3 + y = e^x -x + 9$

a) Show that for every x, there exists only exactly one y.

[snip]
For a given value of x, $e^x - x + 9 = a > 0$. And it's simple to show that $h(y) = y^3 + y - a$ is a one-to one function and that there's exactly one solution to h(y) = 0 for all values of x. Therefore .....

3. Originally Posted by Spec
$y^3 + y = e^x -x + 9$

a) Show that for every x, there exists only exactly one y.

b) Let the solution $y$ in (a) be $y = f(x).$ Show that this function is $C^1$ locally.

c) With the previous two answers in mind, how does it now follow that $f \in C^1(R)$

d) What is the domain and range of f.
Let $g(x) = e^x -x + 9$ and $h(y) = y^3 + y$. Then g and h are both infinitely differentiable functions. Also, h is strictly increasing, with domain and range both equal to R, so by the inverse function theorem h has a continuously differentiable inverse $h^{-1}$. The range of g is the half line [10,∞), with the minimum value 10 occurring at x=0.

If h(y) = g(x) then $y=h^{-1}(g(x))$, which by the chain rule is a continuously differentiable function of x. The domain of this function is the whole of R, and its range is the half line $[h^{-1}(10),\infty)$.

[I can't help wondering whether the person who set this question was thinking that g(0)=9 rather than 10. In that case, since h(2)=9, the answer to the last part of the question would come out much more cleanly: the range of f would then be the half line [2,∞).]