Find the derivative of

e="f(x) = \frac{6}{x} " alt="f(x) = \frac{6}{x} " align="absmiddle" class="latex-formula">
at x = 2.

Mathematics

1 answer:

aev [14]3 years ago

8 0

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

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One equation for this would be

$y = \frac{41}{16} x-\frac{55}{8}$

We start by finding the slope between the two points:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}}
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\\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5})
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A line parallel to this one will have the same slope. We will use point-slope form to write our equation:

$y-y_1=m(x-x_1)
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\\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5})
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\\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5}
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\\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80}
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\\
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The answer to the question