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# calculus problem

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The function $f(x),$ defined for $x > 0,$ is positive, differentiable, and decreasing. The function $f(x)$ has the following property: Let $P$ be a point on the graph, and draw the tangent to the graph at $P.$ Let the tangent meet the $x$-axis and $y$-axis at $A$ and $B,$ respectively.  Then the length of chord $AB$ is contant.

Find all functions $f(x)$ that have this property.

Aug 26, 2023

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Let's analyze the given property. We have a function $$f(x)$$ that is positive, differentiable, and decreasing. When we draw a tangent to the graph of $$f(x)$$ at a point $$(x, f(x))$$, the tangent intersects the x-axis at $$(x - f(x), 0)$$ and the y-axis at $$(0, f(x))$$. This forms a right triangle with the x-axis and y-axis, and the tangent line acts as the hypotenuse of the triangle.

Given that the length of chord $$MN$$ (which is the hypotenuse of the right triangle) is constant, it means that the triangle's hypotenuse has a fixed length. Let's denote this constant length as $$k$$.

Using the Pythagorean theorem for the right triangle, we have:

$(x - f(x))^2 + f(x)^2 = k^2.$

Simplify:

$x^2 - 2xf(x) + f(x)^2 + f(x)^2 = k^2.$

Combine like terms:

$x^2 - 2xf(x) + 2f(x)^2 = k^2.$

Now, we need to find a function $$f(x)$$ that satisfies this equation for all $$x$$ where the function is defined.

Notice that $$k^2$$ is a constant, so the equation implies that $$x^2 - 2xf(x) + 2f(x)^2$$ is also a constant. This means that the derivative of this expression with respect to $$x$$ must be $$0$$:

$\frac{d}{dx} (x^2 - 2xf(x) + 2f(x)^2) = 0.$

Simplify and differentiate:

$2x - 2f(x) - 2xf'(x) + 4f(x)f'(x) = 0.$

Factor out $$2$$ and $$f'(x)$$:

$2(f(x) - x) + 2f'(x)(2f(x) - x) = 0.$

Since $$f(x)$$ is decreasing and positive, $$f'(x) < 0$$, which means that the equation above holds if and only if both terms in the equation are $$0$$:

$f(x) - x = 0$
$2f(x) - x = 0.$

Solve these two equations for $$f(x)$$:

$f(x) = x$
$f(x) = \frac{x}{2}.$

Now, let's check if these solutions satisfy the given conditions.

For $$f(x) = x$$, it's a linear function, which is both decreasing and increasing. This doesn't satisfy the requirement of being only decreasing.

For $$f(x) = \frac{x}{2}$$, it is a decreasing function, and the length of the chord formed by the tangent is indeed constant (equal to $$\frac{k}{2}$$).

So, the only function that satisfies the given property is $$f(x) = \frac{x}{2}$$.

Aug 26, 2023