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I'm trying to do the following exercise:

Find the equation of the line tangent to the graph of $f(x)= (x-1)^{1/3}$ at the point $x=2$. Find also if that line intersects the graph of f(x) at any other point. If it does, then find its coordinates.

Well, I begin by differentiating f(x) to get the slope:

$ f(x) = (x-1)^{1/3}$

$ f'(x) = (1/3)(x-1)^{-2/3}$

$ f'(2) = (1/3)(2-1)^{-2/3}$

$ f'(2) = (1/3)1^{-2/3}$

$ f'(2) = (1/3)1$

$ f'(2) = 1/3$

So the equation of the tangent line at 2 should be $y = (1/3)x$ + constant

To find the constant I use the value of f(2) which is 1.

$f(2) = (2-1)^{1/3}$

$f(2) = 1^{1/3}$

$f(2) = 1$

$f(2) = $ tangent line at 2

$1 = (1/3)2 + constant$

$1 = 2/3 + constant$

$1 - 2/3 = constant$

$1/3 = constant$

Therefore the equation of the tangent line at 2 is $y=(1/3)(x+1)$

Now I try to find out whether there are other points where this line intersects the function.

$(x-1)^{1/3} = (1/3)(x+1)$

$(x-1) = (1/27)(x+1)^3$

$\frac{x-1}{(x+1)^3} = (1/27)$

And here I have no idea of how to continue.

So I cheated a little bit by graphing the function (black curve) and its tangent at 2 (red line):

enter image description here

Looking at the graph, there seems to be another point of intersection at $x=-7$.

Plugging in -7, I get:

$\frac{-7-1}{(-7+1)^3} = (1/27)$

$\frac{-8}{(-6)^3} = (1/27)$

$\frac{-8}{-216} = (1/27)$

$\frac{8}{216} = (1/27)$

This seems to confirm that at $x=-7$ the line intersects the function again. But suppose I wanted to find the point by hand, without graphing and 'guessing' that the answer is -7. How do I do that?

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    $\begingroup$ Just a constructive point for your next question, you can omit some of the trivial lines so that your question isn't un-necessarily long. $\endgroup$ – mrnovice Apr 17 '17 at 22:59
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Your

$$\frac{x-1}{(x+1)^3} = \frac{1}{27}$$

corresponds to

$$x^3+3x^2-24x+28=0$$

but since this comes from a tangent, you know that two of the solutions are $x=2$, so you can factor out $(x-2)^2$ to give

$$(x-2)^2(x+7)=0$$

so the other solution is $x=-7$

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  • $\begingroup$ Could you please elaborate a little more on this? Why are two of the solutions at $x=2$? I don't really understand the idea. $\endgroup$ – IMK Apr 17 '17 at 23:17
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    $\begingroup$ A cubic equation has three solutions, though some might be duplicated and some might be complex. Your expression finds the $x$ coordinates of points where the curve and the tangent meet. If they cross simply, then there would be a solution at that point (as there is with $x=-7$). So you know that $(x-2)$ is a factor. But in fact this is a tangent, which you might informally regard as crossing and crossing back at the same point, leading to two solutions (at least) at that point $\endgroup$ – Henry Apr 18 '17 at 0:02
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HINT: $$\frac{x-1}{(x+1)^3}=\frac{1}{27}$$ can be written as $$f(x)=x^3+3x^2-24x+28$$ We already know that $x-2$ is a factor . Thus we re-write it as: $$f(x)=(x-2)(x^2+5x-14)$$ $$f(x)=(x-2)^2(x-7)$$ Thus it intersects again @ $x=7$

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