# Can a Function Have Two Different Tangent Lines at the Same Point?

• Jaquis2345
In summary: Therefore, f can only have one tangent line at the point (x, f(x)).In summary, the given proof shows that if f is a function and x is in the domain of f, then f cannot have two tangent lines at the point (x, f(x)). This is proven by showing that if there were two tangent lines, then there would be two derivatives of f at the point x that are not equal, which contradicts the definition of a derivative. Thus, f can only have one tangent line at the point (x, f(x)).
Jaquis2345
Homework Statement
If f is a function and x is in the domain of f, then f does not
have two tangent lines at the point (x, f (x)).
Relevant Equations
This is the definition I used for this question. It is a topological definition for the derivative of a function at a point and we are only concerned with the real number line.

Def: The function f has derivative D at the number x in the domain of f if x is a limit point of the domain of f and for every open interval S containing D there is an open interval T containing x such that if t is a
number in T and in the domain of f such that t /= x, then
(f (t)-f (x))/(t-x) is contained by S.
(Note: I just want someone to check my proof and give me advice on anything that seems off. Thank you. )
Proof:
Suppose f is a function and x is in the domain of f s.t. there is a derivative at the point x and sppse. there are two tangent lines at the point (x,f(x)). Let t1 represent one of the tangent lines at (x,f(x)) and let t2 represent the other tangent line at (x,f(x)) s.t. the slopes of t1 and t2 are different. Since the slope of a tangent line is the derivative of the function, the function f would have two derivatives at the point x that are not equal since the slope of t1 is not equal not the slope at t2.
Let D1 denote the slope of the tangent line t1 at (x,f(x)) and let D2 denote the slope of the tangent line t2 at (x,f(x)) s.t. D1< D2. So, D1 and D2 are both derivatives of f at the point x. Let S = (a, b) be the open interval containing D1 and D2. So, a<D1<D2<b.
Let T = (c, d) be the open interval containing x. Since there is a derivative at the x which is in the domain of f, x is a limit point of the domain of f. Thus, there exists a point t in the domain of f s.t. t is contained by T and t /= x. Since D1 and D2 are both derivatives of f at the point x and D1 and D2 are contained by S, [f(t)-f(x)]/(t-x) is contained by S.
Suppose a<D1<[f(t)-f(x)]/(t-x)<D2<b. Since [f(t)-f(x)]/(t-x)<D2 there exists a point w s.t.
w = ([f(t)-f(x)]/(t-x)+D2)/2 (w is the point between these two values. Thus, the open interval (a,w) would contain D1 and [f(t)-f(x)]/(t-x) but not D2. Thus, D2 is not the derivative of f at the point x and t2 is not tangent to the point (x, f(x)). Since a derivative does exist at x, D1 must be the derivative at x. Thus, t1 is the line tangent to (x, f(x)). Similarly, it can be shown that t2 is the only line tangent to (x, f(x)).
Therefore, f can only have one tangent line at the point (x, f(x)).

Jaquis2345 said:
Homework Statement:: If f is a function and x is in the domain of f, then f does not
have two tangent lines at the point (x, f (x)).
Relevant Equations:: This is the definition I used for this question. It is a topological definition for the derivative of a function at a point and we are only concerned with the real number line.

Def: The function f has derivative D at the number x in the domain of f if x is a limit point of the domain of f and for every open interval S containing D there is an open interval T containing x such that if t is a
number in T and in the domain of f such that t /= x, then
(f (t)-f (x))/(t-x) is contained by S.
(Note: I just want someone to check my proof and give me advice on anything that seems off. Thank you. )

Proof:
Suppose f is a function and x is in the domain of f s.t. there is a derivative at the point x and sppse. there are two tangent lines at the point (x,f(x)). Let t1 represent one of the tangent lines at (x,f(x)) and let t2 represent the other tangent line at (x,f(x)) s.t. the slopes of t1 and t2 are different. Since the slope of a tangent line is the derivative of the function, the function f would have two derivatives at the point x that are not equal since the slope of t1 is not equal not the slope at t2.
Let D1 denote the slope of the tangent line t1 at (x,f(x)) and let D2 denote the slope of the tangent line t2 at (x,f(x)) s.t. D1< D2. So, D1 and D2 are both derivatives of f at the point x. Let S = (a, b) be the an open interval containing D1 and D2. So, a<D1<D2<b.
Let T = (c, d) be the an open interval containing x. Since there is a derivative at the x which is in the domain of f, x is a limit point of the domain of f. Thus, there exists a point t in the domain of f s.t. t is contained by T and t /= x. Since D1 and D2 are both derivatives of f at the point x and D1 and D2 are contained by S, [f(t)-f(x)]/(t-x) is contained by S.
Suppose a<D1<[f(t)-f(x)]/(t-x)<D2<b.
Ok. I don't see anything about the two other cases.
Jaquis2345 said:
Since [f(t)-f(x)]/(t-x)<D2 there exists a point w s.t.
w = ([f(t)-f(x)]/(t-x)+D2)/2 (w is the point between these two values. Thus, the open interval (a,w) would contain D1 and [f(t)-f(x)]/(t-x) but not D2. Thus, D2 is not the derivative of f at the point x and t2 is not tangent to the point (x, f(x)). Since a derivative does exist at x, D1 must be the derivative at x. Thus, t1 is the line tangent to (x, f(x)). Similarly, it can be shown that t2 is the only line tangent to (x, f(x)).
Therefore, f can only have one tangent line at the point (x, f(x)).
I suggest that you divide [D1,D2] into three segments, [D1, D1+l/3), [D1+l/3, D1+2*l/3], and (D2-l/3, D2], where l=D2-D1. Find an open interval, T, such that (f(t)-f (x))/(t-x) is in both (D1-I/3, D1+I/3) and (D2-I/3, D2+I/3). Show that this is a contradiction.

Jaquis2345 said:
Homework Statement:: If f is a function and x is in the domain of f, then f does not
have two tangent lines at the point (x, f (x)).
Relevant Equations:: This is the definition I used for this question. It is a topological definition for the derivative of a function at a point and we are only concerned with the real number line.

Def: The function f has derivative D at the number x in the domain of f if x is a limit point of the domain of f and for every open interval S containing D there is an open interval T containing x such that if t is a
number in T and in the domain of f such that t /= x, then
(f (t)-f (x))/(t-x) is contained by S.
(Note: I just want someone to check my proof and give me advice on anything that seems off. Thank you. )
Essentially what you have to prove is that the derivative of ##f## at ##x## is unique. For this, I suggest you should be using the properties of open sets.

The layout of your proof makes it difficult to read, especially as you are not using Latex:

https://www.physicsforums.com/help/latexhelp/

That said, your proof looks painfully muddled to me. Why would you start with an interval ##(a, b)## containing both ##D_1## and ##D_2##? That seems like the opposite of what you want to do.

Instead, you should start with disjoint open intervals ##S_1, S_2## s.t. ##D_1 \in S_1## and ##D_2 \in S_2##.

Note: you may wish to prove such open intervals exist for any two non-equal real numbers, as a preliminary lemma. Although, depending on what you've covered on your course, this may be something you can take as already established.

PS this is, in fact, a key topological property of real numbers.

## 1. What is the purpose of proof in real analysis?

Proof in real analysis is used to establish the validity of mathematical statements and theorems. It is essential for ensuring the accuracy and rigor of mathematical arguments.

## 2. How is proof different in real analysis compared to other branches of mathematics?

In real analysis, proofs often involve the use of limits, continuity, and convergence. These concepts are unique to real analysis and require a different approach compared to other branches of mathematics.

## 3. What are some common techniques used in proof for real analysis?

Some common techniques used in proof for real analysis include direct proof, proof by contradiction, and proof by induction. These techniques involve logical reasoning and use of mathematical definitions and properties.

## 4. How important is rigor in proof for real analysis?

Rigor is crucial in proof for real analysis as it ensures that the conclusions drawn from the proof are accurate and valid. It also allows for a clear understanding of the underlying concepts and their implications.

## 5. Can real analysis proofs be applied to real-world problems?

Yes, real analysis proofs can be applied to real-world problems in fields such as physics, engineering, and economics. The concepts and techniques used in real analysis can help in understanding and solving complex real-world problems.

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