Question

Suppose that a certain television program has an average point rating of 5.8 (ratings measure the number of viewers), and that the program's executives estimate that a 0.1 drop in the ratings will correspond to a $4.4 million drop in revenue. If f(p) is the revenue earned in millions of dollars in terms of p, the ratings points, express this estimation as a derivative.

**answer choices attached inside! please explain :) thanks!!

Answer 1

A,B,C,D,E from top to bottom :)

***i think it's either C or D :/

what do yall think? please explain!!! :)

Answer 2

No, I don't understand this problem.

Answer 3

oh okay:( haha darn :(

Answer 4

why do you think its C or D?

remember derivative = slope = change in y/ change in x

Answer 5

because it matches the values in the problem?

idk really :(

could you please explain it to me from the beginning? I think i'm misunderstanding it :/

Answer 6

ok lets use definition of slope here
the x value we are evaluating at is 5.8, so answer is f'(5.8) = ?

\[f'(5.8) = \frac{f(5.8 -0.1) - f(5.8)}{-0.1} = \frac{-4.4}{-0.1} = 44\]

Answer 7

ahh okay i see now... forgot to apply it to f(a+h)-f(a)/h :/

so that would automatically narrow it down to answer choices A and B right? :/

Answer 8

lol no thats the answer, only 1 possibility

Answer 9

ohh okay so it's answer choice A ?

Answer 10

yep

Answer 11

oh okay awesome!!

i was wondering, for that last problem

Find the equation of the tangent line to the curve y=√x at x=4,

we found the equation (y=1/4 x +1 ) through the power rule...

how would we find it through the

f(a+h)-f(a)
---------
h
equation?

Answer 12

would it be plugging in 4 into the "a" areas to find the slope? :/

Answer 13

just wondering if you knew haha or if it was even possible to find it through that equation :P @dumbcow :) if not, then it's okay haha

Answer 14

haha wait nvm lol :P

thank you for your help!!!

Answer 15

if using the definition of the derivative, you'd find the limit

\[f'(x) = \lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \] To do that, multiply by \[\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}\] and then you can evaluate the limit directly.

Answer 16

so for the other problem i added above, Find the equation of the tangent line to the curve y=√x at x=4,


i'd be doing this?

f'(x) =lim h->0 (sq.rt. 4+h) - sqrt.4
-----------------------
h

and then now what happens? @whpalmer4 :)

Answer 17

No, you have to do it symbolically, find the limit, and only then can you plug in the value. You're finding the function that describes f'(x) first (by taking the limit), and then evaluating it at a point to find the slope of the tangent line to the original f(x) there.

Answer 18

oh darn:(

so what would it look like?

Answer 19

\[f'(x) = \lim_{h\rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h\rightarrow 0}(\frac{\sqrt{x+h}-\sqrt{x}}{h}*\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}})\]\[= \lim_{h\rightarrow 0}\frac{(\sqrt{x}+\sqrt{x+h})(\sqrt{x+h}-\sqrt{x}) }{h(\sqrt{x+h}+\sqrt{x})}\]you finish it off...

Answer 20

okay so from here, you plug in right?

so would you cancel out the (sqrt. x+h) + sq.rt. x ?

which would leave you with lim h->0 (sq.rt. x+h) - sq.rt x
-----------------------
h

?? and then you plug in the 4 into where it says x ?

Answer 21

@whpalmer4 does that seem right to you so far? :/

Answer 22

No... :-( You have to expand the top to get something that will cancel out that h on the bottom. If we just wanted h on the bottom (which prevents us from evaluating the limit directly), we wouldn't have had to do anything, we started out that way :-)

Answer 23

Protip: notice that the numerator is now the product of (a+b)(a-b) for suitable values of a, b.

Answer 24

aww darn okay... ermm would i be multiplying it by something sq.rt x + h ?

Answer 25

so that we would end up with positivex+h ? ;/

not sure.... @whpalmer4 :/

if it's not that, what do i do to cancel out the h at the bottom?

Answer 26

Come on, do the algebra!

What is \[(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})\]That's the numerator of the fraction.

Answer 27

if you prefer, let \(a = \sqrt{x+h}\) and \(b = \sqrt{x}\) and do the algebra like that, then substitute back.

Answer 28

\[(\sqrt{x+h}^2-\sqrt{x}^2)\]

??

Answer 29

did i do that right?? @whpalmer4 :) :/

Answer 30

well, keep going...

Answer 31

so now i plug in x=4?

Answer 32

no, keep going until you've simplified it all the way

Answer 33

(x+h) - x ? @whpalmer4 :)

Answer 34

keep going

Answer 35

x-x =0

so you're left with h?

or do you sub in 4 now:/

Answer 36

you still haven't evaluated the limit. must evaluate the limit to find the function f'(x) before you substitute in any values!

Answer 37

well, actually I suppose you could get away with substituting in the value of x at this point, but there's no reason not to finish the algebra first.

Answer 38

how do i finisht eh algebra :/ i'm kinda confused here :(

Answer 39

\[f'(x) = \lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\]Can you do any more simplificat

Answer 40

simplification?

Answer 41

1/ sq.rt. x+h + sq.rt x ?

Answer 42

\[f'(x) = \lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} =\lim_{h\rightarrow 0}\frac{\cancel{h}}{\cancel{h}(\sqrt{x+h}+\sqrt{x})}\]\[= \lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\]

Answer 43

Now you can evaluate that directly, and now you have my permission to substitute \(x = 4\) :-)

Answer 44

1
--------
2√h + 2 ?

Answer 45

= 1
--------

2√0 + 2

= 1/4 ?

Answer 46

Also, now you see why we use the power rule instead of the definition of the derivative :-)

As h->0, what does that fraction equal?

Answer 47

haha yeah this is way more complicated!! :/

did i do that right?

Answer 48

actually, I didn't call you on it, you can't move the 4 out from inside the radical sign like you did! after h->0, then the 4 is alone under the radical and you can take the square root.

Answer 49

again, had you done the limit in its entirety before messing around with substituting values, you wouldn't have been temped!

\[\lim_{h\rightarrow 0 }\frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{2\sqrt{x}}\]

Answer 50

sigh. tempted.

Answer 51

ohhh

so this is how it should be done?

lim h->0 1/ (sq.rt 4+h) + sq.rt 4 = 1/ (sq.rt. 4 + 0 ) + sq.rt. 4 = 1/ 4 ? @whpalmer4 :)

Answer 52

No. Do the limit first. When you've done that, plug in values. Same amount of work, but you're left with something useful if you need to find the tangent anywhere else on the graph!

\[f'(x) = \lim_{h\rightarrow0}\frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{\sqrt{x+0}+\sqrt{x}} = \frac{1}{2\sqrt{x}}\]Now plug in any value of \(x\) you wish (except 0).

Answer 53

ahh okay i see

and from there

1/2(sq.rt.4) = 1/4 ?

Answer 54

Exactly.

Answer 55

ah okay yay!! awesome!! thank you!! :)

okay, so basically to summarize, you need to use lim h-> 0 f(a+h)-f(a) / h

get rid of the h variables

do the limit (h->0), then sub in the x=4 value?

did i miss anything? haha :p

Answer 56

If you're doing it with the definition of the derivative, then use, you find that limit. You may not have to do anything much to be able to evaluate it directly, or it may require many steps to find. Once you've found the limit, you have an expression which you can evaluate at any point in the domain of the function to find the slope of the tangent line at that point.

Answer 57

ah okay awesome!! Thanks so much!!! :D

Answer 58

not sure what happened to the structure of my first sentence there :-)

Answer 59

ah, got it, I meant to say "yes" not "use" but typed "ues" instead of "yes" and the system "helpfully" corrected it to "use".

Answer 60

hahaha yeah i kinda just thought you said "then you find the limit." lol...computers always think they know what we're trying to say :P haha thanks again!!