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The function is $f(x)=1/x^2$.

I have to find equation of the tangent line that also goes through the point $(0,12)$.

I know how to construct this problem in desmos: https://www.desmos.com/calculator/bj1as3oqo0

So by reading the graph I know solutions are:

  • $y=-16x+12$
  • $y=16x+12$

But how do I find this equations only by calculation? How do I figure out when the line becomes the tangent in this case?

I think I'm pretty good at understanding derivatives and tangent lines, but this one is a hard case for me.

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We have, the equation of any tangent at $x_0$ is: $g(x)=f'(x_0)(x-x_0)+f(x_0)$, $(x_0 \neq 0)$.

We have to find the tangent such as $g(0)=12$ which means $f'(x_0)(-x_0)+f(x_0)=12$ so, $\frac{2}{x_0²}+\frac{1}{x_0²}=12$ By multiplicating both sides by $x_0²$ we find: $3=12x_0²$ Wich means either $x_0=1/2$ or $x_0=-1/2$ .

You can verify by replacing $x_0$ by 1/2 or -1/2 in the equation of the tangent, you will find $y=-16x+12$ and $y=16x+12$

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  • $\begingroup$ Thank you very much! I didn't know about the tangent formula at any x! $\endgroup$ – user119422 Feb 22 '14 at 14:58
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$$f(x)=\frac1{x^2}=x^{-2}\implies f'(x)=-2x^{-3}$$

The parametric equation of $\displaystyle f(x)=\frac1{x^2}$ is $\displaystyle\left(t,\dfrac1{t^2}\right)$

$\displaystyle\implies f'(x)_{\text{ at }\left(t,\dfrac1{t^2}\right)}=-\frac2{t^3}$

So, the equation of the tangent will be $$\frac{y-\dfrac1{t^2}}{x-t}=-\frac2{t^3}\iff 2x+yt^3=3t$$

Now use the fact : this line passes through $(0,12)$ to find $t$

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  • $\begingroup$ Is the use of parametric equation the only way? (I'm in high school) --- I understand that "t=1/2" and when I insert this into $f'(x)$ I get $k=-16$ $\endgroup$ – user119422 Feb 22 '14 at 11:29
  • $\begingroup$ @user119422, Parametrization often eases calculation. Else, you can solve for $a,b$ with two equations $$b=\frac1{a^2}, \frac{12-b}{0-a}=-\frac2{a^2}$$ $\endgroup$ – lab bhattacharjee Feb 22 '14 at 11:38
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Hint: (a naive approach, but possibly the only one) Take the derivative at any point $x_0$ to determine the $y$-intercept of the tangent line at $x_0$. The $y$-intercepts are a function of $x_0$; find the points where that function is equal to $12$.

[Fair warning: I haven't gone through the calculations, so maybe you end up with some nasty polynomial. But without thinking too hard, looks like it should be a quadratic equation in the end.]

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