anonymous
  • anonymous
Precalculus help!! 1. Find the derivative of f(x) = 6 divided by x at x = -2. and 2. Find the derivative of f(x) = 4x + 7 at x = 5. Thank you. Need help solving, will medal.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
I bet you are practicing using the formal definition and not the short cuts right?
anonymous
  • anonymous
Yes @freckles
freckles
  • freckles
you could use either one of these: \[f'(a)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} \text{ or } f'(a)=\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\]

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freckles
  • freckles
they are the same definition just written a little different
freckles
  • freckles
which one are you more comfortable with?
anonymous
  • anonymous
I'm not sure, I'm still confused with both.
freckles
  • freckles
Ok we could do question one together both ways and you can decide which way you want to use for the second and I will check your work for that one
anonymous
  • anonymous
Ok thanks!
freckles
  • freckles
\[a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f(a+h)=f(-2+h)=\frac{6}{-2+h}\] first do you understand how I got f(a+h) and f(a) for the first one?
anonymous
  • anonymous
Yes completely.
freckles
  • freckles
Ok I'm going to input them into the formula
freckles
  • freckles
\[f'(a)=f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}-(-3)}{h}\]
freckles
  • freckles
\[f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}+3}{h}\]
freckles
  • freckles
ok the trick is to get that h on the bottom to cancel somehow so we can actually plug in h=0
freckles
  • freckles
notice we have a compound fraction
freckles
  • freckles
that is a fraction inside a bigger fraction
anonymous
  • anonymous
Hold on super quick, writing this down.
anonymous
  • anonymous
Ok got it.
freckles
  • freckles
we want to see if we can "simplify" the fraction that means get rid of the compound fraction notice the little fraction has denominator (-2+h) multiply top and bottom by (-2+h) this will make our compound fraction into a "simple" fraction do you want to try multiplying top and bottom by (-2+h)
freckles
  • freckles
and let me know what the result is just doing that one step
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Do I multiply the top and bottom of just the little fraction?
freckles
  • freckles
multiply top and bottom by (-2+h) of the big fraction
anonymous
  • anonymous
Would it be 9/h(-2+h) ?
freckles
  • freckles
that isn't a bad try ... I think the 9 is just a bit off...should be 3 there \[\lim_{h \rightarrow 0} \frac{\frac{6}{-2+h}+3}{h} \cdot \frac{-2+h}{-2+h}=?\] \[\lim_{h \rightarrow 0} \frac{\frac{6(-2+h)}{-2+h}+3(-2+h)}{h(-2+h)} \\ =\lim_{h \rightarrow 0}\frac{6+3(-2+h)}{h(-2+h)}\]
freckles
  • freckles
3(-2+h)=-6+3h
freckles
  • freckles
\[\lim_{h \rightarrow 0} \frac{6-6+3h}{h(-2+h)}\]
anonymous
  • anonymous
Oh ok, I see. So it would be 3h/h(-2+h)?
freckles
  • freckles
\[\lim_{h \rightarrow 0}\frac{3h}{h(-2+h)}\] you should recall h/h=1 (we can say this since h is not 0)
freckles
  • freckles
\[\lim_{h \rightarrow 0}\frac{3 \cancel{h}}{\cancel{h}(-2+h)}\]
freckles
  • freckles
guess what ?
anonymous
  • anonymous
So then we are left with 3/-2+h
freckles
  • freckles
we are going to have no problems pluggin in 0 for h now because this function 3/(-2+h) actually exists at h=0
freckles
  • freckles
so what do you think f'(-2)=?
anonymous
  • anonymous
But how do you know that the function exists at 0?
freckles
  • freckles
because the bottom is not 0 when h is 0
anonymous
  • anonymous
Oh ok got it, so would f'(-2) = 3/-2?
freckles
  • freckles
yes
freckles
  • freckles
or -3/2
freckles
  • freckles
doesn't matter where you put the negative at
anonymous
  • anonymous
Could you check my other question once I solve it?
freckles
  • freckles
k.. while you are working it I'm going to do the first one again but use the other formula I wrote
anonymous
  • anonymous
Ok (:
anonymous
  • anonymous
Is the second one f(5) = 4?
freckles
  • freckles
\[f'(a)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} \\ a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f'(a)=f'(-2)=\lim_{x \rightarrow -2} \frac{\frac{6}{x}+3}{x+2} \\ \text{ multiply \top and bottom by } x \\ f'(-2)=\lim_{x \rightarrow -2} \frac{6+3(x)}{(x+2)(x)} \\ f'(-2)=\lim_{x \rightarrow a}\frac{3(x+2)}{(x+2)(x)} \\ \text{ notice} \frac{x+2}{x+2}=1 \text{ for when } x \neq -2 \\ f'(-2)=\lim_{x \rightarrow -2} \frac{3}{x} =\frac{3}{-2} \text{ or } \frac{-3}{2} \text{ or } -\frac{3}{2} \text{ or } -1.5\]
freckles
  • freckles
oops one sec let me check
freckles
  • freckles
"2. Find the derivative of f(x) = 4x + 7 at x = 5." f(5)=4(5)+7=20+7=27
anonymous
  • anonymous
Oh ok, I'm used to the first version we used though. But the one you just used also makes a lot of sense. (:
anonymous
  • anonymous
Yes, and for f(a+h) I got it = to 27+h
freckles
  • freckles
did you mean f'(5)=4?
anonymous
  • anonymous
Yes that's what I meant.
freckles
  • freckles
that is right
anonymous
  • anonymous
Could you check this one too? The position of an object at time t is given by s(t) = -8 - 9t. Find the instantaneous velocity at t = 1 by finding the derivative.
freckles
  • freckles
yeah this question is basically the same as the last questions you asked just asked it in word problem form
anonymous
  • anonymous
Would it be: f'(1) = 9?
freckles
  • freckles
the velocity=derivative of position v(t)=s'(t) They are using to find s'(1)
freckles
  • freckles
close
freckles
  • freckles
not exactly though
freckles
  • freckles
you are missing a sign
freckles
  • freckles
\[v(1)=s'(1)=\lim_{h \rightarrow 0}\frac{s(1+h)-s(1)}{h}\]
anonymous
  • anonymous
-9?
freckles
  • freckles
yes
anonymous
  • anonymous
My bad, didn't catch the negative sign.
freckles
  • freckles
are you using short cuts? or the long way?
freckles
  • freckles
because you found that really fast
anonymous
  • anonymous
The long way but using mental math
freckles
  • freckles
nice
anonymous
  • anonymous
I have the steps written down though, the long way. Would this be ok for an answer: First, I used: s(t) = -8 - 9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. f(a) = -8 - 9t = -8 - 9(1) = -17 f(a+h) = -8 - 9t = -8 - 9(1+h) = -17 - 9h Then I used the equation: f'(a) = lim[f(a+h) - f(a)/h] = f'(a) = -17 - 9h - (-17)/h = -9h/h = -9 Thus I found the derivative to be: f'(1) = -9
freckles
  • freckles
s(t) = -8 - 9t notice the slope of s(t) is -9 the derivative of s(t) is -9
freckles
  • freckles
but yeah your teacher probably wants you to use that formal definition of derivative thing but you know we remember from algebra that the slope of f(x)=mx+b is m and we learn from calculus that the slope is the derivative
anonymous
  • anonymous
Ok, but is the way I showed my work using the formal definition?
freckles
  • freckles
yes it is pretty too
freckles
  • freckles
you did very well
freckles
  • freckles
only thing I would say and it is really not apart of your work "s(t) = -8 - 9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. :" I would say f(a+h) and f(a) but I knew what you meant when you said this
freckles
  • freckles
oops s(a+h) and s(a) since our function name is s not f
freckles
  • freckles
you could change all those f letters to s letters actually
anonymous
  • anonymous
Ok thanks!
freckles
  • freckles
np
anonymous
  • anonymous
Thanks for all your help!!
anonymous
  • anonymous
I will be doing another test similar if you could help me later on? @freckles Maybe in an hour or so.

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