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What is $f'(a)$ of the function $f(x)$?

$$f(x) = 2x^2 − 3x + 1$$

I have been trying to do it but I cannot figure out how to do it. Please somebody can help me.

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    $\begingroup$ Do you know how to calculate $f'(x)$? $\endgroup$ – Arthur Jan 25 '17 at 22:46
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    $\begingroup$ Yes I do know, it is 4x-3 $\endgroup$ – David Romero Jan 25 '17 at 22:47
  • $\begingroup$ Just replace x with a now. $\endgroup$ – randomgirl Jan 25 '17 at 22:49
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    $\begingroup$ $f'(a) \to \left. {\frac{d}{{dx}}f(x)\,} \right|_{\,x\, = \,a} $ $\endgroup$ – G Cab Jan 25 '17 at 22:57
  • $\begingroup$ Related: Finding $f'(a)$ for a given value $\endgroup$ – Jack Jan 26 '17 at 0:37
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When we write $f'(x) = 4x-3$, that means that $x$ is a placeholder. In other words, exactly what symbol we use is of little consequence. Thus, we have, for instance, $$ f'(\color{red}x) = 4\color{red}x-3\\ f'(\color{red}5) = 4\cdot \color{red}5 - 3\\ f'(\color{red}\dagger) = 4\color{red}\dagger - 3\\ f'(\color{red}わ) = 4\color{red}わ - 3 $$ and, of course, $f'(\color{red}a) = 4\color{red}a - 3$.

The (almost) only time to be careful is when we insert specific numbers. See, we might have written $f'(5) = 17$, which is technically correct, but that makes it difficult to see exactly how the $5$ we insert affected the result. However, when using other symbols, like $a, x,$ etc., this is usually not a problem.

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We have $$f'(x) = 4x - 3$$ by simply using that $$(x^n)' = nx^{n-1}$$ for any $n \in \mathbb{N}$. Then you just plug $a$ in for $x$, so $$f'(a) = 4a - 3$$

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