text: # Derivative Problem
  
  • Aug 15th 2009, 02:18 PM
  Derivative Problem
  Suppose $\displaystyle f$ is a function that satisifies the equation $\displaystyle f(x+y)=f(x)+f(y)+yx^2+xy^2$ for all real $\displaystyle x,y$. Suppose that $\displaystyle \lim_{x->0}\frac{f(x)}{x}=1$. Find $\displaystyle f(0),f'(0),f'(x)$.
  
  I star by writing $\displaystyle x+y=0,x=-y, y=-x$
  
  $\displaystyle f(0)=f(x)+f(-x)+x^3-x^3$
  $\displaystyle =f(x)+f(-x)$
  
  So now I have:
  
  $\displaystyle f(x)=f(0)-f(-x)$
  
  $\displaystyle f(0)=f(0)-f(0)=0$
  
  For $\displaystyle f'(0)$:
  
  $\displaystyle f'(0)=\lim_{x->0}\frac{f(x)-f(0)}{x-0}$
  
  $\displaystyle =\lim_{x->0}\frac{f(x)}{x}=1$
  
  So I got $\displaystyle f'(0)=1,f(0)=0$. I can't seem to get started on finding $\displaystyle f'(x)$.
  • Aug 15th 2009, 02:36 PM
  NonCommAlg
  Quote:
  
  Suppose $\displaystyle f$ is a function that satisifies the equation $\displaystyle f(x+y)=f(x)+f(y)+yx^2+xy^2$ for all real $\displaystyle x,y$. Suppose that $\displaystyle \lim_{x->0}\frac{f(x)}{x}=1$. Find $\displaystyle f(0),f'(0),f'(x)$.
  
  I star by writing $\displaystyle x+y=0,x=-y, y=-x$
  
  $\displaystyle f(0)=f(x)+f(-x)+x^3-x^3$
  $\displaystyle =f(x)+f(-x)$
  
  So now I have:
  
  $\displaystyle f(x)=f(0)-f(-x)$
  
  $\displaystyle f(0)=f(0)-f(0)=0$
  
  For $\displaystyle f'(0)$:
  
  $\displaystyle f'(0)=\lim_{x->0}\frac{f(x)-f(0)}{x-0}$
  
  $\displaystyle =\lim_{x->0}\frac{f(x)}{x}=1$
  
  So I got $\displaystyle f'(0)=1,f(0)=0$. I can't seem to get started on finding $\displaystyle f'(x)$.
  
  to get $\displaystyle f(0) = 0$ it's easier to put $\displaystyle y = 0.$ to find $\displaystyle f'(x)$: $\displaystyle f'(x)=\lim_{y \to 0} \frac{f(x+y)-f(x)}{y}=\lim_{y\to0} \left(\frac{f(y)}{y} + x^2 + xy \right)=x^2 + 1.$ this also gives you $\displaystyle f(x)=\frac{x^3}{3} + x.$
explanation: The document contains a math problem and a partial solution to it. The problem is to find \( f(0) \), \( f'(0) \), and \( f'(x) \) given the functional equation \( f(x+y) = f(x) + f(y) + yx^2 + xy^2 \) and the limit condition \( \lim_{x \to 0} \frac{f(x)}{x} = 1 \). The user has provided a partial solution, finding \( f(0) \) and \( f'(0) \), but is stuck on finding \( f'(x) \). The document also includes a complete solution provided by another user.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Derivative Problem
  
  • Aug 15th 2009, 02:18 PM
  Derivative Problem
  Suppose $\displaystyle f$ is a function that satisifies the equation $\displaystyle f(x+y)=f(x)+f(y)+yx^2+xy^2$ for all real $\displaystyle x,y$. Suppose that $\displaystyle \lim_{x->0}\frac{f(x)}{x}=1$. Find $\displaystyle f(0),f'(0),f'(x)$.
  
  I star by writing $\displaystyle x+y=0,x=-y, y=-x$
  
  $\displaystyle f(0)=f(x)+f(-x)+x^3-x^3$
  $\displaystyle =f(x)+f(-x)$
  
  So now I have:
  
  $\displaystyle f(x)=f(0)-f(-x)$
  
  $\displaystyle f(0)=f(0)-f(0)=0$
  
  For $\displaystyle f'(0)$:
  
  $\displaystyle f'(0)=\lim_{x->0}\frac{f(x)-f(0)}{x-0}$
  
  $\displaystyle =\lim_{x->0}\frac{f(x)}{x}=1$
  
  So I got $\displaystyle f'(0)=1,f(0)=0$. I can't seem to get started on finding $\displaystyle f'(x)$.
  • Aug 15th 2009, 02:36 PM
  NonCommAlg
  Quote:
  
  Suppose $\displaystyle f$ is a function that satisifies the equation $\displaystyle f(x+y)=f(x)+f(y)+yx^2+xy^2$ for all real $\displaystyle x,y$. Suppose that $\displaystyle \lim_{x->0}\frac{f(x)}{x}=1$. Find $\displaystyle f(0),f'(0),f'(x)$.
  
  I star by writing $\displaystyle x+y=0,x=-y, y=-x$
  
  $\displaystyle f(0)=f(x)+f(-x)+x^3-x^3$
  $\displaystyle =f(x)+f(-x)$
  
  So now I have:
  
  $\displaystyle f(x)=f(0)-f(-x)$
  
  $\displaystyle f(0)=f(0)-f(0)=0$
  
  For $\displaystyle f'(0)$:
  
  $\displaystyle f'(0)=\lim_{x->0}\frac{f(x)-f(0)}{x-0}$
  
  $\displaystyle =\lim_{x->0}\frac{f(x)}{x}=1$
  
  So I got $\displaystyle f'(0)=1,f(0)=0$. I can't seem to get started on finding $\displaystyle f'(x)$.
  
  to get $\displaystyle f(0) = 0$ it's easier to put $\displaystyle y = 0.$ to find $\displaystyle f'(x)$: $\displaystyle f'(x)=\lim_{y \to 0} \frac{f(x+y)-f(x)}{y}=\lim_{y\to0} \left(\frac{f(y)}{y} + x^2 + xy \right)=x^2 + 1.$ this also gives you $\displaystyle f(x)=\frac{x^3}{3} + x.$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math problem and a partial solution to it. The problem is to find \( f(0) \), \( f'(0) \), and \( f'(x) \) given the functional equation \( f(x+y) = f(x) + f(y) + yx^2 + xy^2 \) and the limit condition \( \lim_{x \to 0} \frac{f(x)}{x} = 1 \). The user has provided a partial solution, finding \( f(0) \) and \( f'(0) \), but is stuck on finding \( f'(x) \). The document also includes a complete solution provided by another user.
  
  Label: question-and-answer