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The question says that $y = kx - 2$ is a tangent to $3x^2 + x + 1$ and I have to find $k$. It's pretty easy, you can equate the two equations.

However, how do we use differential calculus to do it:

So, the first differential: $f'(x) := 6x + 1$

then we equate it: $kx - 2 = 6x + 1$ and get: $\frac{3}{x} + 6$. What to do now?

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3 Answers 3

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Notice that $f'(a)$ is the slope of the tangent line to the curve at $(x,y)=(a,f(a))$. So, you are looking for $a$ such that $6a+1=k$.

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If $y=kx-2$ is tangent to the curve $y=3x^2+x+1\;,$ Then These two equation has exactly one

Solution. So Put $y=kx-2$ in $y=3x^2+x+1\;,$ We get $3x^2+x+1 = kx-2$

We get $3x^2+(1-k)x+3=0\;,$ Now for exactly one solution $\bf{Discriminant =0}$

So $$(1-k)^2-4\cdot 3 \cdot 3 = 0\Rightarrow (k-1)^2-(6)^2=0$$

So we get $$(k-1-6)\cdot (k-1+6) = 0\Rightarrow k=7\;\;,k=-5$$

$$\bf{Using \; calculus::}$$

If $f(x)=kx-2$ and $g(x)=3x^2+x+1$ are tangent to each other, Then Slope of tangent

at there common points $P(x_{1},y_{1})$ are equal.

So $$\displaystyle \left[f'(x)\right]_{(x_{1},y_{1})} = k$$ and $$\displaystyle \left[g'(x)\right]_{(x_{1},y_{1})} = 6x_{1}+1$$

So $$\displaystyle \left[f'(x)\right]_{(x_{1},y_{1})} = \left[g'(x)\right]_{(x_{1},y_{1})}\Rightarrow k=6x_{1}+1$$

Now Points $P(x_{1},y_{1})$ also lie on $f(x)$ and $g(x)$ (bcz it is a common point)

So $$kx_{1}-2 = 3x^2_{1}+x_{1}+1\Rightarrow (6x_{1}+1)x_{1}-2 = 3x^2_{1}+x_{1}+1$$

so we get $$3x^2_{1} = 3\Rightarrow x_{1} = \pm 1$$

Now Put into $k=6x_{1}+1\;,$ We get $k=7$ and $k=-5$

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  • $\begingroup$ I know how to use that method; I was asking about the approach using calculus. $\endgroup$
    – weirdpanda
    Sep 8, 2015 at 15:30
  • $\begingroup$ Ah, that is what I was looking for. A little question, what is that square bracket notation signifying? And aren't slope and tangents the same? I mean, that's what we have studied till now. I may be wrong. $\endgroup$
    – weirdpanda
    Sep 8, 2015 at 16:24
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You have to find the equation of the tangent line passing through an arbitrary point $(x',y')$. That equation is $$y - y' = m(x-x') + b$$ or equivalently $$y = mx + y' - mx',$$ where $m$ can be found by that differential. Can you take it from here?

UPDATE Note that $y' = 3x'^2+3x'+1$ and $m = 6x'+3$ so you end up with $$ y = mx + y' - mx' = (6x'+3)x + 3x'^2+3x'+1 - (6x'+3)x' = (6x'+3)x - 3x'^2+1 $$ and since you know the equation is $y=kx-2$, we must have $$ \left. \begin{cases} k &= 6x'+3\\ -2 &= 1-3x'^2 \end{cases} \right\| $$ Can you take it from here?

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  • $\begingroup$ That's all the question says; I had the equation till that point, but after that, I have no clue what to do. :( $\endgroup$
    – weirdpanda
    Sep 8, 2015 at 15:33
  • $\begingroup$ @weirdpanda please see the update, can you finish this? the second equation is straight-forward $\endgroup$
    – gt6989b
    Sep 8, 2015 at 15:59
  • $\begingroup$ Perfect, thanks a lot! :) $\endgroup$
    – weirdpanda
    Sep 8, 2015 at 16:22

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