# Thread: Puzzling to me

1. ## Puzzling to me

Yeah ok I know this is probably simple but damn it all for some reason I can't get my head to figure it out and it has upset me finally. Well here it is.

Draw a graph of a continuous function y=f(x) that satisfies all of these conditions: f'(x) > 0 for x < -2, f'(x) < 0 for -2 < x < 2, and f'(x) = 0 for x > 2.

A little calc for you all. I'm sure you've missed it. oh and all those are f prime (x) when you get to the greater signs in case it's hard to read.

2. ## Re: Puzzling to me

The first two sub-domains :
f'(x) > 0 for x < -2
f'(x) < 0 for -2 < x < 2

could be satisfied by many functions with a turning point at x = -2. The simplest would be a parabola of the form:
f{x) = a(x + 2)^2 + k with a < 0

The third sub-domain:
f'(x) = 0 for x > 2
is satisfied by any function of the form f(x) = constant. If you want the function to be continuous over the whole domain, then:
f(x) = 16a + k

So one (composite) function family to satisfy the constraints is:
f{x) = a(x + 2)^2 + k (a < 0) x <= 2
f(x) = 16a + k x >= 2

Alan

3. ## Re: Puzzling to me

Note: I didn't read the question carefully enough. Sorry. Please see Alan Miller's reply.

The derivative <code>f'</code> must be positive, then negative, then positive again. Think parabola, think quadratic function. The quadratic function must be zero at x= -2 and x=+2, so for example <code><big>f'(x)=x

4. ## Re: Puzzling to me

Attached: modified example; function is continuous, and its derivative is continuous too.

5. ## Re: Puzzling to me

Ah yes, Hans. I hadn't added the (assumed) constraint of a "smooth" curve. I similarly furthered my analysis to come up with a function family of the type:

f(x) = ax( (x^2)/3 - 4 ) + C for x <= 2
f(x) = -16a/3 + C for x >= 2

I think yours belongs to this.

Alan

6. ## Re: Puzzling to me

The original problem was to create a continuous function. When I read your reply and saw that my original reply was incorrect, it was easy to modify it to yield a continuously differentiable function, but that was not required, however.

My "new" solution is indeed part of yours, with a = 3, C = 0. And of course, there are many other solutions, like three linear functions:

f(x) = a(x+2) + c for x less than or equal to -2, with a > 0, c arbitrary
f(x) = b(x+2) + c for x between -2 and 2, with b < 0
f(x) = 4b + c for x greater than or equal to 2

This one is not differentiable at both x=-2 and x=2.

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