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I'm reading Spivak's Calculus, there's a part where he suggests that the student should check some assertions:

$$f(x)=x^2$$

Then I've evaluated for $f(x+1)$ which is $f(x+1)=(x+1)^2=x^2+2x+1$. Why he says that $f(x+1)=f(x)+2x+1$? Does $f(x)=x^2$ in this case?

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    Because $f(x)=x^2$.2012-12-15
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    It is also $f(x+1)=f(x)+f^\prime(x)+1$. :)2012-12-15
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    @AndreaMori Why does $f'(x)=2x$?2012-12-15
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    @GustavoBandeira : if $f(x)=x^2$, what would $f^\prime(x)$ be?2012-12-15
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    Yes, but you created this premise now, it wasn't stated anywhere.2012-12-15
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    $f'(x)$ is the derivative of $f(x)$. If you know $f(x)=x^2$, then $f'(x)=2x$. This isn't an independent premise, it comes directly from what Clayton said.2012-12-15
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    @RobertMastragostino Now it's clear. I still don't know about derivatives. Invisible premises everywhere.2012-12-15
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    @GustavoBandeira : $f(x)=x^2$ is the starting point of your question ...2012-12-15
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    @Andrea Robert explained what I didn't know.2012-12-15

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Since $f(x)=x^2$, $$f(x+1)=(x+1)^2=x^2+2x+1=f(x)+2x+1$$ Both are correct

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    Got it. But this notation left me a little confused.2012-12-15
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    @GustavoBandeira I see. Glad to help2012-12-15
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    @GustavoBandeira: I've not found out what is the problem trying to say. Of course, the answwer is complete.+12012-12-15
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    @BabakSorouh My assumption is that it's testing the algebra needed to set up $\frac{f(x+h)-f(x)}h$, and prepping the student to recognize the cancellation that will occur.2012-12-15