For Prose Posted March 10, 2014 Share Posted March 10, 2014 Hello everyone. I don't want the answer to this question, only for someone to help point me in the right direction. The question is as follows: Suppose that f is differentiable on I:= (-inf,1), such that f(x) > 0 and f'(x) = f(x)^2 for all x belonging to I. Find f knowing that f(0) = 1. Thanks for the hints and help in advance. Link to comment Share on other sites More sharing options...

John Posted March 10, 2014 Share Posted March 10, 2014 (edited) What class is this?As for a hint, [math]f'(x) = f(x)^{2}[/math] is a separable differential equation. If the class doesn't assume knowledge of differential equations, then we may want to try another avenue. Edited March 10, 2014 by John Link to comment Share on other sites More sharing options...

studiot Posted March 10, 2014 Share Posted March 10, 2014 (edited) A hint for an alternative method. You have shown an open interval [math]\left( { - \infty ,1} \right)[/math] What happens when x=1 and when x>1 ie what could be the reason is f(x) not differentiable there? You have the y intercept. Does f(x) ever cross the x axis? Edited March 10, 2014 by studiot Link to comment Share on other sites More sharing options...

For Prose Posted March 10, 2014 Author Share Posted March 10, 2014 What class is this? As for a hint, [math]f'(x) = f(x)^{2}[/math] is a separable differential equation. If the class doesn't assume knowledge of differential equations, then we may want to try another avenue. It is for calculus w/ analytical geometry John. Unless differential equations is commonly referred to by some other name, I don't believe I have gotten there quite yet. Link to comment Share on other sites More sharing options...

studiot Posted March 10, 2014 Share Posted March 10, 2014 (edited) It is for calculus w/ analytical geometry John. Unless differential equations is commonly referred to by some other name, I don't believe I have gotten there quite yet. So what did you make of my alternative suggestions? What have you deduced about f(x), given f'(x) is a quadratic? We don't do your homework for you, just try to offer helpful pointers in the right direction. Edited March 10, 2014 by studiot Link to comment Share on other sites More sharing options...

For Prose Posted March 10, 2014 Author Share Posted March 10, 2014 So what did you make of my alternative suggestions? What have you deduced about f(x), given f'(x) is a quadratic? We don't do your homework for you, just try to offer helpful pointers in the right direction. Hello everyone. I don't want the answer to this question, only for someone to help point me in the right direction. The question is as follows: Suppose that f is differentiable on I:= (-inf,1), such that f(x) > 0 and f'(x) = f(x)^2 for all x belonging to I. Find f knowing that f(0) = 1. Thanks for the hints and help in advance. I was only answering his question, not disregarding yours. If the derivative of f(x) equals f(x)^2, does this mean that to find the original equation for the function, I will have to reverse differatiate (I think this is called integration in Calc 2 but I am nowhere near utilizing this yet)? What have you deduced about f(x), given f'(x) is a quadratic?To be honest, I am not sure what it means that f' is quadratic. I can only reason that you are referring to how it might be differentiated. Thanks for your time involved in this matter studiot. Link to comment Share on other sites More sharing options...

studiot Posted March 10, 2014 Share Posted March 10, 2014 (edited) I'm assuming you know that f'(x) is the derived function or derivative and that that means? A quadratic has the form ax^{2}+bx+c. What function, when differentiated, leads to this quadratic? Note that you can obtain superscript (for the square) and subscript by using the X^{2} and X_{2} icons, just to the side of bold, underline etc. Yes what you must do is officially called integration, but it you can do this without a fancy name. If the derivative of bx^{2 }is 2bx, then the function that, when differentiated, leads to 2bx is bx^{2}. Now am not quite sure what you mean by f'(x) = f(x)^{2, }perhaps we should clear that up, but I am assuming we can identify the constants a, b and c from this. Do you mean [math]\frac{{dy}}{{dx}} = {y^2}[/math] or [math]\frac{{dy}}{{dx}} = a{x^2}[/math] Edited March 10, 2014 by studiot Link to comment Share on other sites More sharing options...

John Posted March 10, 2014 Share Posted March 10, 2014 (edited) If the derivative of f(x) equals f(x)^2, does this mean that to find the original equation for the function, I will have to reverse differatiate (I think this is called integration in Calc 2 but I am nowhere near utilizing this yet)? The process is called antidifferentiation, and solving the differential equation would involve doing just that. However, in an introductory calculus class, it's probably better to just consider the derivatives you've already learned, and think about what sort of function has a derivative equal to its square (or at least, close to its square, since you'll have some decisions to make about certain terms). As an example, consider f(x) = 0. Then f'(x) = 0 = 0^{2} = f(x)^{2}. However, this doesn't work, since the exercise requires that f(0) = 1. Although the fact that f is differentiable on (-∞, 1) doesn't necessarily mean it's not differentiable on [1, ∞), the fact that they specify that interval is suggestive as well. Edited March 10, 2014 by John Link to comment Share on other sites More sharing options...

studiot Posted March 10, 2014 Share Posted March 10, 2014 (edited) and think about what sort of function has a derivative equal to its square (or at least, close to its square, since you'll have some decisions to make about certain terms). As an example, consider f(x) = 0. Then f'(x) = 0 = 0^{2} = f(x)^{2}. However, this doesn't work, since the exercise requires that f(0) = 1. If we do mean Then should we not be writing f'(x) = {f(x)}^{2} That is why I asked my earlier question. Edited March 10, 2014 by studiot Link to comment Share on other sites More sharing options...

John Posted March 11, 2014 Share Posted March 11, 2014 (edited) The added braces might be nice for clarity, I suppose. If we take f'(x) = f(x)^{2} to indicate that f' is a quadratic function, then we can no longer determine f uniquely from the information given. Also, polynomials are closed under integration, so the choice of the text authors to mention the interval (-∞, 1) seems odd, since if f is a polynomial, then it's differentiable on the entirety of (-∞, ∞). That is, of course, unless we take f and f' to be piecewise-defined, but that seems a bit convoluted for a problem I'm assuming is intended to help solidify the reader's knowledge of common derivatives and understanding of function transformations and translations. Edited March 11, 2014 by John Link to comment Share on other sites More sharing options...

For Prose Posted March 11, 2014 Author Share Posted March 11, 2014 When the question states to find f, is f referring to the equation of f? Or is f defined as being f of x? For added clarity, I will post a picture of the question in its original form when I get home. Link to comment Share on other sites More sharing options...

John Posted March 11, 2014 Share Posted March 11, 2014 The notation is sometimes used interchangeably, but f is the function itself, while f(x) is the value of f when evaluated at x. So if we consider f(x) = x^{4} - 6, where x is a real number, then what we have is a function f : ℝ → ℝ defined by f(x) = x^{4} - 6. The question is asking us what formula defines f if f has these various properties. I guess the answer to both parts of your question is "yes," i.e. we're asked to find the equation defining f, and f will be defined by f(x). Link to comment Share on other sites More sharing options...

For Prose Posted March 12, 2014 Author Share Posted March 12, 2014 It seems that a viable solution would be to figure out an equation that fits all these parameters through trial and error. We are currently learning the chain rule. Link to comment Share on other sites More sharing options...

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