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February 1st, 2018, 04:04 PM   #1
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From: MI

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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 Senior Member   Joined: Sep 2016 From: USA Posts: 379 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics 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.
 February 1st, 2018, 05:13 PM #3 Newbie   Joined: Jan 2018 From: MI Posts: 10 Thanks: 1 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.
 February 1st, 2018, 07:01 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,962 Thanks: 1606 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. Thanks from ProofOfALifetime
 February 2nd, 2018, 08:27 AM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 2,751 Thanks: 1401 $\displaystyle f'(a) = \lim_{x \to a} \dfrac{f(x)-f(a)}{x - a}$ $\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{f(x)-f(-10)}{x - (-10)}$ $\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{8x^2+6x+11 - 751}{x +10}$ $\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{8x^2+6x-740}{x +10}$ $\displaystyle f'(-10) = \lim_{x \to -10} \dfrac{2(4x-37)(\cancel{x+10})}{\cancel{x +10}} = 2(-77) = -154$ Thanks from greg1313

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