MHB Tangent equation for two points.

[Petrus]

Hello MHB,
I am doing some old exam and got less knowledge for this problem.
consider the graph of the function
$$g(x)=\frac{1}{x}, \ x\neq 0$$
the point $$\left(3,-1 \right)$$ lies on two tangents to the graph. Decide tangents equation.

My progress
well I derivate and find the slope
$$g'(x)=\ln(x)$$ so the slope is $$m=\ln(3)$$
so the tangents equation is $$y=\ln(3)(x-3)-1$$ is this correct? I am supposed to get two tangent equation

Regards,
$$|\pi\rangle$$

 

[alyafey22]

well I differentiated and find the slope
$$g'(x)=\ln(x)$$ so the slope is $$m=\ln(3)$$

I think you got confused . Try to differentiate again .

 

[Petrus]

I think you got confused . Try to differentiate again .

I have done to much integrate today...
I mean $$-\frac{1}{x^2}$$ and we get $$y=-\frac{1}{9}(x-3)-1$$ But is this correct method? How shall I find second tangent point

Regards,
$$|\pi\rangle$$

 

[alyafey22]

If $f$ is differentiable at a point $p$ , can it have more than one tangent line at $p$ ?

 

[Opalg]

I have done to much integrate today...
I mean $$-\frac{1}{x^2}$$ and we get $$y=-\frac{1}{9}(x-3)-1$$ But is this correct method? How shall I find second tangent point

The point $(3,-1)$ does not lie on the curve, so it does not make sense to look for the tangent at that point. What you need to do is to find the equation of the tangent at a general point on the curve, say the point $(p, 1/p)$, and then apply the condition for that tangent to go through the point $(3,-1)$.

 

[Petrus]

Zaid,
I actually honestly not 100% sure if it can or not, I keep thinking but I don't really know and interested if it can.

The point $(3,-1)$ does not lie on the curve, so it does not make sense to look for the tangent at that point. What you need to do is to find the equation of the tangent at a general point on the curve, say the point $(p, 1/p)$, and then apply the condition for that tangent to go through the point $(3,-1)$.

Hello Opalg,
You are obviously correct.. I did not even think about that if it is in the point $$\frac{1}{3} \neq -1$$
I get the tangent equation for $$(p, 1/p)$$
$$y=-\frac{1}{p^2}(x-p)+\frac{1}{p}$$
replace x and y with the point etc we get $$a_1=1 \ a_2=-3$$
so the two tangent line is $$y=-x+2, \ y=-\frac{x}{9}-\frac{2}{3} $$
that agree with the facit
Thanks for the help and fest respond from you both !
Regards,
$$|\pi\rangle$$

 

[MarkFL]

Here is a pre-calculus technique:

Let the tangent line be:

$$y=mx+b$$

Now, equate this to the function:

$$mx+b=\frac{1}{x}$$

$$mx^2+bx-1=0$$

Since the line is tangent, there will be 1 repeated root, hence the discriminant is zero:

$$b^2+4m=0$$

$$m=-\left(\frac{b}{2} \right)^2$$

Hence, the tangent line is:

$$y=-\left(\frac{b}{2} \right)^2x+b$$

We are told it must pass through the point $$(3,-1)$$ so we find:

$$-1=-\left(\frac{b}{2} \right)^2(3)+b$$

Thus:

$$3b^2-4b-4=0$$

$$(3b+2)(b-2)=0$$

For:

i) $$b=-\frac{2}{3}\,\therefore\,m=\frac{1}{9}$$

The tangent line for this root is:

$$y=-\frac{1}{9}x-\frac{2}{3}$$

ii) $$b=2,\therefore\,m=-1$$

The tangent line for this root is:

$$y=-x+2$$

 

[alyafey22]

I actually honestly not 100% sure if it can or not, I keep thinking but I don't really know and interested if it can.

It is very important to know that a differentiable function at a point $$a$$ can't have more than one tangent line . This is basically what we mean by $$f'(x)$$ exists . Now, suppose you have a function that has a discontinuity at a point ,say, $$b$$ then we cannot draw a tangent line at this point . So , continuity is necessary for differentiability . But , suppose that a function is continuous on an open interval can we conclude it is differentiable at any point in the interval ?

Remember that a differentiable function at a point means that we can draw '' a unique tangent line '' at this point .

 

[alyafey22]

I have done to much integrate today...

By the way . Solving lots of problems will not enhance ''understanding the concept'' .