Is it possible for 2 tangent lines of this function to be perpendicular?

[unf0r5ak3n]

Homework Statement

Is it possible that there are two different tangent lines to this function that are perpendicular to each other? If so, find the equations of the two lines and show that they are tangent to ƒ(x) and perpendicular to each other. If not, show why it is not possible.

Homework Equations

ƒ(x) = x^2 + 2

The Attempt at a Solution

I believe it is possible, I found the line tangent to ƒ(x) at x=2 and its perpendicular line was fairly close to a tangent to the function. However I'm not sure how to find the equations that would be exactly tangent to ƒ(x) as well as perpendicular to one another

 

[Quinzio]

Start writing a general equation for a tangent of f(x)

 

[unf0r5ak3n]

f(2) = 4x-2

 

[Quinzio]

A general equation f'(x) = 2x

 

[unf0r5ak3n]

right the derivative is 2x

 

[Quinzio]

Ok, so what can you say about the equations of 2 perpendicular lines ?
Equation like y= mx+q

 

[unf0r5ak3n]

they should be reciprocals

 

[Mark44]

they should be reciprocals

The equations won't be reciprocals.

 

[unf0r5ak3n]

?? confused

 

[Quinzio]

they should be reciprocals

Be reciprocal is one change, the other change ?

You're given y= 3x+1

Write a perpendicular line.
y = ?

 

[unf0r5ak3n]

y = (x-1)/3

 

[Mark44]

?? confused

The slopes of the two lines will be reciprocals.

 

[unf0r5ak3n]

so no 2 tangents can form a perpendicular line at f(x) = x^2+2?

 

[Quinzio]

so no 2 tangents can form a perpendicular line at f(x) = x^2+2?

Yes they can, and they do in an infinite number of points.
But you have to learn to write equations of perpendicular lines before.

 

[unf0r5ak3n]

what would be an example and how do you find the initial one?

 

[Mark44]

what would be an example and how do you find the initial one?

An example of what? The initial one what? Please try to write sentences that are clear.

 

[unf0r5ak3n]

What would be an example of two equations of lines that are perpendicular and tangent to the function f(x)=x^2+2?
Would any line form a perpendicular angle as well as be tangent to the function? If not, how do you find the equation of the first line as appose to the second one?

 

[jgm340]

TO THE OP: I think you may be misunderstanding the question you were asked. The question is: find two tangent lines to f(x) such that the two are perpendicular to each other. In other words, they will both be parallel to the function f(x) at their respective points, but they will be perpendicular to each other.

********************

Now, you already know that the derivative of f(x) = x^2+2 is f'(x) = 2x. THAT is the slope of the tangent line of f(x) at x.

The question can be rephrased as: Find two points x₁ and x₂ such that f'(x₁) is the negative reciprocal of f'(x₂).

Using the fact that f'(x) = 2x, this is equivalent to asking: Find two points x₁ and x₂ such that 2x₁ is the negative reciprocal of 2x₂. Can you find two such points?

 

[unf0r5ak3n]

like f(1)=2 and f(-.25)=-.5?

 

[unf0r5ak3n]

f'(x) for both*

 

[jgm340]

Yes, exactly!

 

[unf0r5ak3n]

ok, but i don't understand what to use them for?

 

[jgm340]

Well, f'(1) is the slope of the tangent line to f(x) at x = 1, and f'(-0.25) is the slope of the tangent line to f(x) at x = -0.25.

So, you are looking for two lines:

y = a₁x + b₁

and

y = a₂x + b₂

That are tangent to f(x) at x=x₁ and x = x₂, respectively, but which are perpendicular to each other.

You have just found the slopes that these tangent lines need to have! That is, you have found that a₁ should equal 2, and a₂ should equal -0.5. This automatically makes the two lines you are creating perpendicular to each other. Now you just need to choose b₁ and b₂ so that the first line intersects the point (x₁, f(x₁)) and the second line intersects the point (x₂,f(x₂)).

 

[unf0r5ak3n]

oh! i actually figured it out as you posted that last one comment, thank you so much!

curious if this is actually correct,
In the function f(x)=x^2+2, lines y=2x+1 and y=.5x+1.9375 are tangent to f(x) and are also perpendicular.

 

[Mark44]

oh! i actually figured it out as you posted that last one comment, thank you so much!

curious if this is actually correct,
In the function f(x)=x^2+2, lines y=2x+1 and y=.5x+1.9375 are tangent to f(x) and are also perpendicular.

These lines aren't perpendicular. The slope of the first line is 2 and that of the second line is 1/2. These are reciprocals, but not the negative reciprocals of each other.

Also, the lines have to be tangent to the curve y = x^2 at points on this curve. I don't believe that the line y = 2x + 1 is actually tangent to this curve anywhere.

 

[jgm340]

Your constants are wrong. Also, that .5x should be a -.5x.

You want to find a constant such that 2*x + b₁ = x^2 + 2 for x = 2.
You also want to find a constant such that -0.5*x + b₂ = x^2 + 2 for x = -0.25