Question

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.

Answer 1

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

Answer 2

Yes @freckles

Answer 3

you could use either one of these:
\[f'(a)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} \text{ or } f'(a)=\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\]

Answer 4

they are the same definition just written a little different

Answer 5

which one are you more comfortable with?

Answer 6

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

Answer 7

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

Answer 8

Ok thanks!

Answer 9

\[a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f(a+h)=f(-2+h)=\frac{6}{-2+h}\]
first do you understand how I got f(a+h) and f(a) for the first one?

Answer 10

Yes completely.

Answer 11

Ok I'm going to input them into the formula

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

notice we have a compound fraction

Answer 16

that is a fraction inside a bigger fraction

Answer 17

Hold on super quick, writing this down.

Answer 18

Ok got it.

Answer 19

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)

Answer 20

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

Answer 21

Yes

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

that isn't a bad try ... I think the 9 is just a bit off...should be 3 there
\[\lim_{h \rightarrow 0} \frac{\frac{6}{-2+h}+3}{h} \cdot \frac{-2+h}{-2+h}=?\]

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

\[\lim_{h \rightarrow 0}\frac{3h}{h(-2+h)}\]
you should recall h/h=1 (we can say this since h is not 0)

Answer 30

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

Answer 31

guess what ?

Answer 32

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

Answer 33

we are going to have no problems pluggin in 0 for h now
because this function 3/(-2+h) actually exists at h=0

Answer 34

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

Answer 35

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

Answer 36

because the bottom is not 0 when h is 0

Answer 37

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

Answer 38

yes

Answer 39

or -3/2

Answer 40

doesn't matter where you put the negative at

Answer 41

Could you check my other question once I solve it?

Answer 42

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

Answer 43

Ok (:

Answer 44

Is the second one f(5) = 4?

Answer 45

\[f'(a)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} \\ a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f'(a)=f'(-2)=\lim_{x \rightarrow -2} \frac{\frac{6}{x}+3}{x+2} \\ \text{ multiply \top and bottom by } x \\ f'(-2)=\lim_{x \rightarrow -2} \frac{6+3(x)}{(x+2)(x)} \\ f'(-2)=\lim_{x \rightarrow a}\frac{3(x+2)}{(x+2)(x)} \\ \text{ notice} \frac{x+2}{x+2}=1 \text{ for when } x \neq -2 \\ f'(-2)=\lim_{x \rightarrow -2} \frac{3}{x} =\frac{3}{-2} \text{ or } \frac{-3}{2} \text{ or } -\frac{3}{2} \text{ or } -1.5\]

Answer 46

oops one sec let me check

Answer 47

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

Answer 48

Oh ok, I'm used to the first version we used though. But the one you just used also makes a lot of sense. (:

Answer 49

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

Answer 50

did you mean f'(5)=4?

Answer 51

Yes that's what I meant.

Answer 52

that is right

Answer 53

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.

Answer 54

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

Answer 55

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

Answer 56

the velocity=derivative of position
v(t)=s'(t)

They are using to find s'(1)

Answer 57

close

Answer 58

not exactly though

Answer 59

you are missing a sign

Answer 60

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

Answer 61

-9?

Answer 62

yes

Answer 63

My bad, didn't catch the negative sign.

Answer 64

are you using short cuts?
or the long way?

Answer 65

because you found that really fast

Answer 66

The long way but using mental math

Answer 67

nice

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

yes it is pretty too

Answer 73

you did very well

Answer 74

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

Answer 75

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

Answer 76

you could change all those f letters to s letters actually

Answer 77

Ok thanks!

Answer 78

np

Answer 79

Thanks for all your help!!

Answer 80

I will be doing another test similar if you could help me later on? @freckles Maybe in an hour or so.