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February 1st, 2018, 04:04 PM   #1
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Calc help

I am not sure what the expression is asking for, I've tried a whole bunch of different solutions but I just don't get it.
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February 1st, 2018, 04:08 PM   #2
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The derivative of a function at $-10$ is the following number if it exists:
\[ \lim\limits_{h \to 0} \frac{f(-10 + h) - f(-10))}{h} .\]
Evaluate this in your example.
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February 1st, 2018, 05:13 PM   #3
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I did that and got 8h-154. I'm not sure what I'm supposed to do next.

Last edited by skipjack; February 1st, 2018 at 06:48 PM.
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February 1st, 2018, 07:01 PM   #4
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You should instead evaluate $\displaystyle \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

and you initially get $\displaystyle \frac{f(x + h) - f(x)}{h} = 8h + 16x + 6$

which has a limit of $16x + 6$ as $h$ tends to 0.

That is the expression that is asked for, and is the derivative of $f(x)$.

Its limit as $x$ tends to -10 can be found by substituting -10 for $x$.

Note: it's a rather poor question.
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February 2nd, 2018, 08:27 AM   #5
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$\displaystyle f'(a) = \lim_{x \to a} \dfrac{f(x)-f(a)}{x - a}$


$\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{f(x)-f(-10)}{x - (-10)}$

$\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{8x^2+6x+11 - 751}{x +10}$

$\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{8x^2+6x-740}{x +10}$

$\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{2(4x-37)(\cancel{x+10})}{\cancel{x +10}} = 2(-77) = -154$
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