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Given the equation $f(x)=2 (x+1)^2$, use the definition $\dfrac{df}{dx}=\displaystyle\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ of derivative to find the slope of the tangent to the graph $f(x)$ at point $(x_1, y_1)$; I have reached the following answer,

$y-2x_1^2-2+4x-4x_1$

I really, really tried to solve but I´m not certain about the if the answer is right.

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    What is the question?2012-07-04

3 Answers 3

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Use the point-slope formula.

However, to get the slope using the limit formula we have $f'(x_1)=\lim_{h\rightarrow 0}\frac{4x_1h+4h+2h^2}{h}=4(x_1+1)$.

So the formula for the tangent line is $y-y_1=4(x_1+1)(x-x_1)$. Since $y_1=2(x_1+1)^2$ then the tangent line equation simplifies to $y=4(x_1+1)x+2(1-x_1^2)$

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    @ViniciusL.Beserra : It's not the same as yours, because yours is not an equation. There's no "$=$" in yours.2012-07-04
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If $f(x)=2(x+1)^2$ then $ \lim_{h\to0}\frac{f(x_1+h)-f(x_1)}{h} = \lim_{h\to0}\frac{2(x_1+h+1)^2-2(x_1+1)^2}{h}. $ The $2$ is a constant and may therefore be pulled out of the limit. (And remember: in this context, "constant" means not depending on $h$.) A bit of trivial algebra---expanding the two squares---says this is equal to $ 2\lim_{h\to0}\frac{(x_1^2+2x_1h+2x_1+h^2+2h+1)-(x_1^2+2x_1+1)}{h} $ Cancelations in the numerator reduce this to $ 2\lim_{h\to0}\frac{2x_1h+h^2+2h}{h}. $ Then this becomes $ 2\lim_{h\to0}\frac{h(2x_1+h+2)}{h} = 2\lim_{h\to0}(2x_1+h+2) = 2(2x_1+2) = 4x_1+4. $

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    In order to facilitate the calculus I first developed (x+1)^2 first finding. 2x^2+2x+2, and after 2(2x^2+2x+2), I used the denifition to find the answer.2012-07-04
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Point-slope form for the tangent line will immediately yield $y - y_1 = f'(x_1) (x - x_1)$ as the tangent line. If you've found the derivative correctly, $f'(x) = 4x - 4$, and so $f'(x_1) = 4x_1-4$.

Think you can fix any errors from there?

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    Yes, i have done and i just reached this same answer of your altough not put it in the question. My doubt is how to find the equation of the slope. So, now for it´s clear that I have to use y1 in the original equation 2(x+1)^2 to find the y1.2012-07-04