Mirrored Functions

Calculus Level 2

Find the shortest distance between the two curves $f(x)=\ln(x)$ and $g(x)=e^x$ .

2\sqrt{2}

2

\frac{1}{\sqrt{2}}

\sqrt{2}

2 solutions

Akhil Bansal
Mar 16, 2016

$f(x) = ln x$ and $g(x) = e^x$ are mirror image of each other about $y = x$
Therefore, for minimum distance , $f'(x) = g'(x) = 1$
$f'(x) = \dfrac{1}{x} = 1 \Rightarrow x = 1 \ ; \ y = 0$ $g'(x) = e^x = 1 \Rightarrow x = 0 \ ; \ y = 1$
Distance between $(1,0)$ and $(0,1)$ is $\sqrt{1^2 + 1^2} = \boxed{\sqrt{2}}$

Moderator note:

There is no justification for this statement:

Therefore, for minimum distance , $f'(x) = g'(x) = 1$

"for minimum distance , $f'(x) = g'(x) = 1$ " Could you please explain this?

Anandhu Raj - 5 years, 3 months ago

Both the curves will be closest to each other when tangent to both of them will be parallel to the line $y = x$ .

This strikes me after plotting the curves on a paper.

Akhil Bansal - 5 years, 3 months ago

That claim is not true, since in particular, the $x$ values that you calculated, are distinct. What you likely mean is that "If the shortest line segment is $(a, f(a)), (b, g(b) )$ , then we have $f'(a) = 1, g'(b) = 1$ .

Even with that, the statement is not true. for example, consider the lines $f(x) = 100x + 1$ and $g(x) = 100x$ . The shortest line segment between them doesn't satisfy $f'(a) = 1, g'(b) = 1$ .

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – Yes correct.. my statement is true for inverse functions only.

Akhil Bansal - 5 years, 3 months ago

@Akhil Bansal – That's still not true as yet. For example, the functions $y = 10x$ and $y = \frac{1}{10} x$ are inverses of each other. However, the minimum distance between them doesn't occur where the gradient is 1.

Think about what is happening, and what the actual mathematically correct statement should be.

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – New Statement : The minimum distance between inverse functions would be the length of common normal to both the curves and also y = x.

This is my original statement. Is it correct now?

Akhil Bansal - 5 years, 2 months ago

@Akhil Bansal – That is not your original statement. Your original statement was $f'(x) = g'(x)$ , where you used the same value of $x$ . That is my main issue with the statement.

If your original intention is indeed for $f'(a) = g'(b)$ and $f^{-1} (a) = b$ , why do you think it is true? Is that a sufficient condition? Is that a necessary condition?

Calvin Lin Staff - 5 years, 2 months ago

Christina Becker
Mar 18, 2016

After drawing both curves and y=x, I noticed that an isoceles right triangle is formed by half the segment between a given point on e^x and its mirror, the vertical line down from the given point, and the line y=x, with the vertical line as the hypotenuse. So by taking e^x - x, dividing by root 2 and doubling the result, I got an expression for the distance between mirror points.

Taking the derivative of this expression and setting it equal to 0, I got x=0 as a minimum, and substituting this into my expression gave a distance of root 2.

Moderator note:

This solution makes the assumption that the minimum distance must occur between a point and it's inverse.

Why must the minimum distance occur between a point and it's inverse?

Calvin Lin Staff - 5 years, 2 months ago

Because the minimum distance of a point from a line occurs where the distance is drawn at a right angle to the line. Since a point and its inverse are reflections across the line y=x, their respective minimum distances from y=x will intersect y=x at the same point. Thus since the three points are all on the same line, min distance from A to B, plus the min distance from B to C, will equal the min distance from A to C.

Christina Becker - 5 years, 2 months ago

You started off with the assumption that "Let's consider a point and its inverse", as opposed to "a point and any other inverse point", which is what we want to study. So, your focus has only been on these special sets of points, and under this restriction it is clear that the minimium must also satisfy this restriction.

As an explicit counterexample, suppose that we were looking at the function $f = \{ (1,0), (1.1, 1) \}$ . Then, picking the point $(1,0)$ , the minimum distance to a point on the inverse graph would be to the point $(1, 1.1)$ (with distance 1.1), instead of to it's inverse point $(0,1)$ (with distance $\sqrt{2}$ .

But otherwise, you are close to a correct argument. It is indeed true that the minimum distance between a graph and it's inverse occurs at a point and it's inverse.

Calvin Lin Staff - 5 years, 2 months ago

Mirrored Functions

2 solutions

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