Algebraic solution to: Do the functions $y=\frac{1}{x}$ and $y=x^3$ ever have the same slope? The exercise doesn't specify how it must be answered, so I chose a graphical proof because I couldn't come up with an algebraic one. Sketching the graphs of $y=\frac{1}{x}$ and $y=x^3$, I noticed that $y=x^3$ always has a nonnegative slope, whereas $y=\frac{1}{x}$ always has a negative slope. Therefore these two functions never have the same slope.
However, I'm wondering if there's a algebraic way of showing this. I thought of differentiating each function and setting the values equal, but I think this would only prove that the two functions don't have the same slope at any particular x, and not that they don't have the same slopes anywhere over their domains.
 A: You can prove the exact observation you took from your sketch (note that sketch $\ne $ proof), and indeed you are on the right track with differentiation - no wonder if the problem is about slopes.
The derivative of $x\mapsto \frac 1x$ is $-\frac 1{x^2}$ (where $x\ne 0$), hence $<0$ as the square of a nonzero number is strictly positive.
The derivative of $x\mapsto x^3$ is $3x^2$, hence $\ge 0$ (again because squares are nonnegative).
A number $<0$ can never equal a number $\ge 0$.
A: By "slope of a function", I assume you mean the slope of the tangent line at a point. ie. the derivative.
So you are looking at $f_1(x) = x^3$ and $f_2(x) = 1/x$, and you want to solve the equation
$$
3x^2 = f_1'(x) = f_2'(x) = \frac{-1}{x^2}
$$
Which gives
$$
x^4 = \frac{-1}{3}
$$
which does not have any real solutions. Is this the kind of proof you were looking for?
A: Let $f(x)=x^3$ and $g(x)=x^{-1}$. We have $f'(x)=3x^2$ and $g'(x)=-x^{-2}$, now we do exactly what you said:
$Im(f')\cap Im(g')=\left\{t\in\mathbb{R}|t\geq0 \right\}\cap\left\{t\in\mathbb{R}|t<0 \right\}=\left\{t\in\mathbb{R}|t\geq0\wedge t<0 \right\}=\emptyset$
