I am just stuck and cannot see how to solve this question, I've have a complete mind blank.
Find the slope of the curve $$y= 2x^3 − 8x^2+1$$
at the point $(2, -15).$
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1Related : http://math.stackexchange.com/questions/460989/slope-of-line-tangent-to-a-curve-at-a-given-point-using-first-principles – lab bhattacharjee Aug 22 '15 at 14:52
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Do you know how to compute the derivative of a function? Do you know what it represents? – tst Aug 22 '15 at 15:04
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Nope I have completely forgotten everything sadly. – user263629 Aug 22 '15 at 15:09
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and yeah that related link helps but I dont know what to do with the +1 at the end – user263629 Aug 22 '15 at 15:13
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The +1 is a constant and it disappears, if you differentiate. – callculus42 Aug 22 '15 at 15:18
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Tthe slope at the point $x$ is given by: $$y'(x)=6x^2-16x$$ Here you can substitute $x=2$.
wythagoras
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Dr. Sonnhard Graubner
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The slope of $ y =2x^3-8x^2+1$ can be found by evaluating $\frac{dy}{dx}$ at $(2, -15)$.
$$ \frac{dy}{dx} = 6x^2-16x$$ When $x=2$, $\text{Slope at (2, -15)} = \frac{dy}{dx} = 6(2^2)-16(2) = ?$
$\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$
Can you do the rest?
John_dydx
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I'd highly appreciate it if you could put the whole solution up as I need it for revision. but nonetheless thanks a lot for your reply – user263629 Aug 22 '15 at 15:21
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Are you sure you can't find $(6 \times 2^2) - (16 \times 2)$? How old are you? – John_dydx Aug 22 '15 at 15:23
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Notice, we have the equation of the curve $$y=2x^3-8x^2+1$$
the slope of the tangent at general point $(x, y)$ of the curve is given as $$\frac{dy}{dx}=\frac{d}{dx}(2x^3-8x^2+1)$$ $$\frac{dy}{dx}=6x^2-16x$$ Hence, the slope of (the tangent) at the given point $(2, -15)$ is given as $$\left[\frac{dy}{dx}\right]_{(2, -15)}=6(2)^2-16(2)=24-32=-8$$
Harish Chandra Rajpoot
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$\frac{d}{dx}$ is a linear operator, so you can apply it to each $x^n$ term in the individual summands, also:
$$\frac{d}{dx}x^n = nx^{n-1}$$
hence:
$$ \begin{align} y&= 2x^3 − 8x^2+1\\ &= 2x^3 − 8x^2+1\color{green}{x^0}\\ \frac{d}{dx}y &= \frac{d}{dx}\left(2x^3 − 8x^2+1\color{green}{x^0}\right)\\ &=2\frac{d}{dx}\left(x^3\right) − 8\frac{d}{dx}\left(x^2\right)+1\color{green}{\frac{d}{dx}\left(x^0\right)}\\ &=2\cdot 3x^2 − 8\cdot 2x\color{green}{^1}+1\color{green}{\cdot 0x^{-1}}\\ &=6x^2-16x\\ \end{align}$$
The parts in green are usually not written out like that. There are usually rules like "a constant summand (1 in this case) becomes 0". But the general rule actually holds for $n=0$, so why bother coming up with additional rules? Math is complicated enough already.
Now insert the value 2:
$$6\cdot2^2-16\cdot 2= 24-32=-8\\$$
What's also good to know is if the point they are asking for is actually part of the function:
$$2\cdot2^3 − 8\cdot2^2+1 = 2\cdot8 - 8\cdot4+1 = 16-32+1=-15\\$$
Great, this wasn't a trick question. The slope is $-8$ at that point.
null
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