Find the tangent between two points

Question

Let $f(x) = x^3$. Show that for every point $P = (x_0, y_0) \in\mathbb R^2$ there is a tangent to the graph of $f(x)$ that passes through $P$.

I said in order for this to be true $f'(x)$ needs to be defined for all $f(x)$ and for all $x$'s. If this is true then we will have a value for all $f'(x)$. This means that $f(x)$ will be able to give us all types of straight lines that both grows and decreases. So I differentiated $f(x)$ to $3x^2$ and here we can see that $f'(x)$ is defined for all x's and thus there is a line for every $(x_0, y_0)$ that passes through the point $p$.

But in the answers they used mean value theorem etc, but I was wondering if my explanation was totally wrong because I got $0$ point on it.

Answer 1

Considering

$$ \cases{ y = x^3\\ y = y_0+m(x-x_0) } $$

after substitution (on intersection) we have

$$ x^3 = y_0+m(x-x_0) $$

but we need tangency so $m$ should be such that

$$ \phi_m(x)=x^3-(y_0+m(x-x_0))=(x-a)(x-b)^2 $$

meaning that $\phi_m(x)$ have a double root and as a consequence

$$ \cases{a b^2+m x_0-y_0=0\\ 2 a b+b^2+m=0\\ a+2 b=0} $$

and after the elimination of $\{a,b\}$

$$ 4m^3-27x_0^2m^2+54x_0y_0m-27y_0^2=0 $$

which means that there exists always a real root $m^*$ satisfying this polynomial.

NOTE

From $$ \cases{2 a b+b^2+m=0\\ a+2 b=0}\Rightarrow \cases{b^2=\frac m3\\ b^3 =\frac{m x_0-y_0}{2}}\Rightarrow \left(\frac m3\right)^3 = \left(\frac{m x_0-y_0}{2}\right)^2 $$

Answer 2

Your explanation is intuitive, and could be plausible with accompanying illustrations, but it isn't rigorous or convincing. If one of my students had answered that question that way, I would have given some credit for understanding the statement.

Every point on the graph of $f(x) = x^3$ will already have a tangent line through it, because the function is differentiable. If $P = (x_0,y_0)$ is not on the the graph, and $(x,x^3)$ is on the graph, then the line between them has slope $\frac{x^3-y_0}{x-x_0}$. That line is tangent to the graph of $f$ if the slope equals $f'(x) = 3x^2$.

So you are looking for a real number $x$ such that $$ \frac{x^3-y_0}{x-x_0} = 3x^2 \iff x^3 - y_0 = 3x^2(x-x_0) \iff 2x^3 - 3x^2 x_0 + y_0 = 0 $$ Every cubic polynomial has a real root, so we know some $x$ satisfies the equation.