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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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    But the question as phrased above was _not_ "Find the equation of the tangent line", but rather "Find the slope of the tangent line."2012-07-04
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    This answer is wrong. It correctly finds the equation of the tangent line, but that was not the question. The question was how to use the definition of derivative to find the _slope_ (the _slope_, not the _equation_) of the tangent line. This answer _asserts_ that the limit as $h\to0$ is $4(x_1+1)$, but the question is how to _show_ that that is the limit.2012-07-04
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    Thanks for your answer, you are right.2012-07-04
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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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    The function was given as $f(x)=2(x+1)^2$, so the derivative is $f'(x)=4x+4$. Besides, the question was how to use the definition of derivative to find this. You haven't done that nor even attempted it.2012-07-04
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    The function was written differently before your edit $(2(x-1)^2)$, and I wrote exactly what was necessary to deduce the correct answer given what little he had written.2012-07-04
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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