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.

- anonymous

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- schrodinger

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- freckles

I bet you are practicing using the formal definition and not the short cuts right?

- anonymous

Yes @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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## More answers

- freckles

they are the same definition just written a little different

- freckles

which one are you more comfortable with?

- anonymous

I'm not sure, I'm still confused with both.

- 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

Ok thanks!

- 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

Yes completely.

- freckles

Ok I'm going to input them into the formula

- freckles

\[f'(a)=f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}-(-3)}{h}\]

- freckles

\[f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}+3}{h}\]

- freckles

ok the trick is to get that h on the bottom to cancel somehow so we can actually plug in h=0

- freckles

notice we have a compound fraction

- freckles

that is a fraction inside a bigger fraction

- anonymous

Hold on super quick, writing this down.

- anonymous

Ok got it.

- 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

and let me know what the result is just doing that one step

- anonymous

Yes

- anonymous

Do I multiply the top and bottom of just the little fraction?

- freckles

multiply top and bottom by (-2+h) of the big fraction

- anonymous

Would it be 9/h(-2+h) ?

- 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

3(-2+h)=-6+3h

- freckles

\[\lim_{h \rightarrow 0} \frac{6-6+3h}{h(-2+h)}\]

- anonymous

Oh ok, I see. So it would be 3h/h(-2+h)?

- 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

\[\lim_{h \rightarrow 0}\frac{3 \cancel{h}}{\cancel{h}(-2+h)}\]

- freckles

guess what ?

- anonymous

So then we are left with 3/-2+h

- 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

so what do you think f'(-2)=?

- anonymous

But how do you know that the function exists at 0?

- freckles

because the bottom is not 0 when h is 0

- anonymous

Oh ok got it, so would f'(-2) = 3/-2?

- freckles

yes

- freckles

or -3/2

- freckles

doesn't matter where you put the negative at

- anonymous

Could you check my other question once I solve it?

- freckles

k..
while you are working it
I'm going to do the first one again but use the other formula I wrote

- anonymous

Ok (:

- anonymous

Is the second one f(5) = 4?

- 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

oops one sec let me check

- freckles

"2. Find the derivative of f(x) = 4x + 7 at x = 5."
f(5)=4(5)+7=20+7=27

- 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

Yes, and for f(a+h) I got it = to 27+h

- freckles

did you mean f'(5)=4?

- anonymous

Yes that's what I meant.

- freckles

that is right

- 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

yeah this question is basically the same as the last questions you asked
just asked it in word problem form

- anonymous

Would it be:
f'(1) = 9?

- freckles

the velocity=derivative of position
v(t)=s'(t)
They are using to find s'(1)

- freckles

close

- freckles

not exactly though

- freckles

you are missing a sign

- freckles

\[v(1)=s'(1)=\lim_{h \rightarrow 0}\frac{s(1+h)-s(1)}{h}\]

- anonymous

-9?

- freckles

yes

- anonymous

My bad, didn't catch the negative sign.

- freckles

are you using short cuts?
or the long way?

- freckles

because you found that really fast

- anonymous

The long way but using mental math

- freckles

nice

- 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

s(t) = -8 - 9t
notice the slope of s(t) is -9
the derivative of s(t) is -9

- 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

Ok, but is the way I showed my work using the formal definition?

- freckles

yes it is pretty too

- freckles

you did very well

- 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

oops s(a+h) and s(a)
since our function name is s not f

- freckles

you could change all those f letters to s letters actually

- anonymous

Ok thanks!

- freckles

np

- anonymous

Thanks for all your help!!

- 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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