Show that the root of $x^{1/3}=1-x$ lies between $0$ and $1$ using the intermediate value theorem I am currently doing a problem that asks me to use the intermediate value theorem to show that $x^{1/3}=1-x$ lies between $0$ and $1$. I want to start by evaluating the function at $0$ and $1$, but it seems the function is undefined because if you plug in $1$, you get $1=0$. This doesn't seem right. Is my interpretation correct?
The problem comes from Stewart's Calculus section 2.5, q 48.
 A: Hint: Let $f(x)=x^{1/3}-1+x$. Now $f(0)=-1$ and $f(1)=1$.
A: Or: at $0$ LHS $= 0 < 1 =$ RHS, while at $1$, LHS $=1 > 0 =$ RHS, so at some $x$ between $0$ and $1$ the two sides have to be equal. 
A: The idea of solving equations with functions goes back to Descartes; instead of solving $f(x)=g(x)$, you can define
$$
F(x)=f(x)-g(x)
$$
and find the points where $F$ vanishes. In your case,
$$
F(x)=\sqrt[3]{x}+x-1
$$
Now
$$
F'(x)=\frac{1}{3\sqrt[3]{x^2}}+1
$$
is not defined for $x=0$ and it is strictly positive for $x\ne0$. Therefore $F$ is strictly increasing (note $F$ is continuous also at $0$).
Since $F(0)=-1$ and $F(1)=1$, we know that $F$ must vanish at a unique point in $(0,1)$.

You can also consider the equivalent equation
$$
x=(1-x)^3
$$
that translates to the function
$$
G(x)=x^3-3x^2+4x-1
$$
The derivative is
$$
G'(x)=3x^2-6x+4
$$
which is everywhere positive (the polynomial has no real roots). So $G$ is increasing and, since $G(0)=-1$ and $G(1)=1$, we have the same conclusion as before.
