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I have:

$f(a + b) = f(a) + f(b) + ab$

$\lim_{h \to 0} \frac{f(h)}{h}\ = 7$

So because $f'(0) = \lim_{h \to x} \frac{f(h)- f(x)}{h}\ = 7$

So that shows that $f(x) = 0$ right? But not really sure how to proceed, Played around with $f(a + -a) = f(a) + f(-a) + -a^2 = 0 $ but still stuck.

Simply Beautiful Art
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JohnDoe
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  • If $f(x)=0$ everywhere, then $f'(x)=0\ne7$. – Simply Beautiful Art Sep 17 '17 at 12:00
  • Taken from How to ask a good question. "Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." I have [edited your post] to reflect this principle. – Simply Beautiful Art Sep 17 '17 at 12:02

1 Answers1

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There holds $$ \frac{f(x+h)-f(x)}{h}=\frac{f(x)+f(h)+xh-f(x)}{h}=\frac{f(h)}{h}+x $$ Taking the limit as $h \to 0$ yields. $$ f'(x)=7+x $$

F. Conrad
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  • Ah that makes it so clear, my solution to another exercise was very close to this. Should have seen it myself. Thanks! – JohnDoe Sep 17 '17 at 12:15