0
$\begingroup$

I have the function: $$f(x)=x^3,\forall x\in\mathbb{R}$$ Now, I want to prove that every tangent of $C_f$ (graphical representation of $f$) in a random point $A(a,f(a))$ different from $O(0,0)$ has another same point $B(b,f(b))$ with $C_f$ which slope is $4$ times bigger than the slope of $C_f$ at $A(a,f(a))$, meaning $f'(b)=4\cdot f'(a)$. Any ideas?

$\endgroup$
3
  • $\begingroup$ Can you obtain $f'(a)$ and $f'(b)$? $\endgroup$ Jan 8, 2018 at 13:20
  • $\begingroup$ no, just random points $\endgroup$ Jan 8, 2018 at 13:25
  • $\begingroup$ What is your definition of a "random" point? Like an arbitrary point? $\endgroup$ Jan 8, 2018 at 13:27

3 Answers 3

3
$\begingroup$

The slope of the tangent at $(a,f(a))$ is $f'(a)=3a^2$. So the equation of the tangent is $$y-a^3=3a^2(x-a)$$. Now suppose this line meets the curve again at $(b,f(b))$, with $b\neq a$, then $$\begin{align}&b^3=a^3+3a^2(b-a)\\\implies&b^3-a^3=3a^2(b-a)\\\implies&b^2+ab+a^2=3a^2\\\implies&(b-a)(b+2a)=0\\\implies&b=-2a\end{align} $$ Now the slope of the curve at $x=-2a$ is $3(-2a)^2=4\cdot f'(a)$.

$\endgroup$
1
  • $\begingroup$ @Arthur I added the details. $\endgroup$
    – QED
    Jan 8, 2018 at 14:17
2
$\begingroup$

We have our $a$, our tangent line $C_{f, a}$, which has slope $3a^2$ and goes through $A = (a, a^3)$, which means that the equation for the line $C_{f, a}$ is $$ y = 3a^2x -2a^3 $$ Now, assuming $a\neq 0$, we're after the second point where the tangent crosses the graph. In other words, we're after the second value of $x$, besides $a$, where the value of the function, and the $y$-coordinate of the line are the same: $$ x^3 = 3a^2x-2a^3\\ x^3-3a^2x+2a^3 = 0 $$ Now, we know that $x = a$ is one solution to this equation. That means that since the right-hand side is $0$, the left-hand side has $x -a$ as a factor. We try to factor it out, and get $$ x^3-3a^2x + 3a^2 = x^3-ax^2 + ax^2-3a^2x+2a^3\\ = x^2(x-a) + ax^2-3a^2x+2a^3\\ = x^2(x-a) + ax^2-a^2x - 2a^2x + 2a^3\\ = x^2(x-a) + ax(x-a)-2a^2(x-a)\\ = (x^2+ax-2a^2)(x-a) $$ so our equation has become $$ (x^2+ax-2a^2)(x-a) = 0 $$ Actually, since we know that the line is tangent to $f$ at $x = a$ (and not just intersecting), we can actually automatically know that the second degree factor still has $x = a$ as a solution (we can also see that $a^2+a\cdot a-2a^2 = 0$). We can factor that out again: $$ (x^2+ax-2a^2)(x-a) = (x^2-ax+ax+ax-2a^2)(x-a)\\ = (x(x-a) + 2a(x-a))(x-a)\\ = (x+2a)(x-a)^2 $$ So our equation now says $$ (x+2a)(x-a)^2 = 0 $$ and we can read the solutions right off: they are $x = a$ (which is the one we knew about) and $x = -2a$. Thus $B = (-2a, (-2a)^3)$ is the second point that the tangent intersects with the graph. The slope at $x = -2a$ is $3(-2a)^2 = 12a^2$, which we can see is exactly four times the slope at $A$, which was $3a^2$.

Note that we didn't really use the assumption $a \neq 0$ during our algebraic manipulations (there were no $\frac1a$ or anything). That's because if $a = 0$, then the second point of intersection is also $x = 0$ (in fact, that one point becaomses a "triple" intersection, since any tangent is a "double intersection" already, which you can see from the factor $x-a$ appearing twice in the factored expression). The slope there is $0$, and $4\cdot 0$ is $0$, so the statement is is still true, in some sense.

$\endgroup$
1
$\begingroup$

Suppose $a,b \in \mathbb{R}$ are points so that $f'(b) = 4f'(a)$, then $$3b^2 = 12a^2 \implies b = \pm2a$$

$$f'(b)=f'(\pm2a)=3(4a^2)=4(3a^2)=4f'(a)$$

$\endgroup$
2
  • $\begingroup$ You've confirmed that if $f'(b) = 4f'(a)$, then $b = \pm 2a$ and $f'(b) = 4f'(a)$? How does this answer the question? How do you know that this is the point that the tangent line intersects? (By the way, it's probably a typo: your last $f'(a)$ should be $4f'(a)$) $\endgroup$
    – Arthur
    Jan 8, 2018 at 13:43
  • $\begingroup$ @Arthur I didn't read the question carefully enough to interpret it in that way, sorry you're right $\endgroup$ Jan 8, 2018 at 16:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .