A Quadratic and its Derivative
My friend, Alexander Karabegov, sent me one of his puzzles. I love the mixture of algebra and calculus.
Puzzle. Describe a real quadratic function f such that the graph of its derivative f′ is tangent to the graph of f.
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math noob:
If the quadratic function is f(x)=ax^2+bx+c (a unequal 0), then f'(x)=2ax+b. Since the derivative is tangential to the quadratic function, the equation f(x)=f'(x) has only one solution. This equation is ax^2+(b-2a)x+(c-b)=0. The discriminant is set to 0, which gives us b^2+4a(a-c)=0. Let’s set b=0. Since a unequal 0, a=c is a must. And it is true: all quadratic functions of the form f(x)=ax^2+a satisfy the condition described in the exercise. Note, that if we don’t set b=0, from the b^2+4a(a-c)=0 equation infinitely many additional (a,b,c) triples could be derived, for example: (a,b,c)=(-1,2,-2)
14 October 2023, 7:26 amAndreas:
Nice puzzle, it seems tricky at first glance until you start thinking about it.
For a quadratic function f(x)=ax^2 + bx + c, the derivative is f'(x) = 2ax + b.
Varying c moves f up and down without affecting f’, so we want to find c that gives only one single solution.
They meet where f(x)=f'(x), or x^2 + (b-2a)x/a + (c-b)/a = 0, which has exactly one solution when c=a + b^2/(4a).
So the answer is: ax^2 + bx + a + b^2/4a
14 October 2023, 1:24 pmZarunias:
The word “describe” is very vague, the most elegant I can come up with is this: The y coordinate of the vertex has to be the same as the “opening” (the value a from its equation y=ax²+bx+c (is there a better name for this value a?))
17 October 2023, 5:09 pmtanyakh:
I was vague intentionally hoping for different ways to describe such functions. Alexander Karabegov’s description is: the imaginary parts of its roots are 1 and -1.
17 October 2023, 5:35 pmZarunias:
I like his description even more.
18 October 2023, 7:40 am