Question

Let
f(x)=(x−5)2
. Find the average rate of change of
f(x)
with respect to
x
from
x= −3
to
x= −3 + h
if
h≠0
. Simplify your answer as much as possible.
Let
f(x)=(x−5)2
. Find the average rate of change of
f(x)
with respect to
x
from
x= −3
to
x= −3 + h
if
h≠0
. Simplify your answer as much as possible.
@Mathematics

Answer 1

i get h or 1. im sure im doing it wrong though.

Answer 2

i also got -16 + h / -67
which is also wrong

Answer 3

You have to differentiate f(x)

Answer 4

That doesnt help me much...

Answer 5

I'm assuming you mean: f(x) = (x−5)^2?

Answer 6

yes. i just copied and pasted.

Answer 7

I assume you could find the average rate of change by finding the rate of change at f(-3) and f(-3+h) and then add those two together, and divide the result by 2 to get the average

Answer 8

and the rate of change (sometimes called acceleration) is found by differentiation

Answer 9

I know how to solve it. I must be doing something wrong though because Im getting the wrong answers.

Answer 10

How do you do the differentiation?

Answer 11

i got 64-16h+h^2-64 / -3+h-3

Answer 12

then i canceled things out

Answer 13

f(x) = (x-5)^2 = x^2 - 10x + 25
f'(x) = 2x - 10
f'(-3) = 2(-3) - 10 = -16
f'(-3+h) = 2(-3+h) - 10 = -6 + 2h - 10
(f'(-3) + f'(-3+h))/2 = (-16 - 6 + 2h - 10)/2 = (-32 + 2h)/2 = -16 + h

Answer 14

I'm not entirely sure this is the right approach, there might be another way to calculate the average. Are you able to check if it's the correct answer?

Answer 15

It's the right answer according to my online hw website. I'm trying to figure out how you got it though because I did not solve it like that.

Answer 16

Hehe good. Where does it go wrong for you?

Answer 17

i got -16+h/-6 earlier. that was the closest i got to the right answer.

Answer 18

the -6 would have went away if it was -3 plus 3 not -3 minus 3

Answer 19

so thats why i had the -6 too

Answer 20

If you want to, you can show me the steps you go through. Because I can't really see how you get that result

Answer 21

hold on a sec

Answer 22

how did u get 2x-10?

Answer 23

Just to be clear, do you know what differentiation is?

Answer 24

ummm kind of

Answer 25

I think you'll need a pretty good understanding of that, to solve these kinds of questions. Here's the wikipedia article http://en.wikipedia.org/wiki/Differentiation_(mathematics). It explains what it is better than I can.

Answer 26

Basically it's a way to find the rate of change in regards to x, if that makes sense

Answer 27

yeah thats what i thought... but i thought you find the change by subtracting old from new on the top and bottom of the fraction and then dividing the fraction and/or canceling things out.

Answer 28

Yes that would work if it was a linear equation, because then the rate of change would be constant.

Answer 29

ok thanks. i think i got it now. haha

Answer 30

So, a = (y2-y2)/(x2-x1) is for linear equations only hehe

Answer 31

ups (y2-y1)/(x2-x1)*

Answer 32

yeah...

Answer 33

o wait u never answered me how u got 2x-10

Answer 34

hehe all right

Answer 35

f(x) = (x-5)^2 = x^2 - 10x + 25, so now we can differentiate f(x). The notation for which is f'(x) (notice the ' sign between the f and the (x) )

Answer 36

(There are other notations that your teacher might prefer, like dx/dy instead of f'(x)

Answer 37

so to differentiate x^2 - 10x + 25 we write f'(x) = (x^2 - 10x + 25)' =

Answer 38

And here we use rules for differentiating

Answer 39

a^x differentiated becomes x*a^(x-1)

Answer 40

so x^2 becomes 2*x^(2-1)

Answer 41

which is 2x^1 = 2x

Answer 42

o wow. i dont recall doing this in class at all.

Answer 43

hehe i hope i'm not trying to teach you things you don't need to understand yet

Answer 44

what are you going through in class at the moment?

Answer 45

We're on Functions part C

Answer 46

hmm that doesn't tell me much

Answer 47

We just started doing rate of change Friday

Answer 48

haha judging by this link http://www.ies.co.jp/math/java/calc/heihen/heihen.html, i think i just made it worse for you

Answer 49

none of your links are working

Answer 50

oh sorry, there's a comma at the end of the link, there you go http://www.ies.co.jp/math/java/calc/heihen/heihen.html

Answer 51

i think you were on the right track

Answer 52

it's basically (f(-3+h)-f(-3))/h

Answer 53

i didn't realize that since you're only interesting in finding the average rate of change you can consider it a linear equation

Answer 54

forgive me for misleading you, you will probably have to learn the other way of doing it, later on though

Answer 55

it's still a = (y2-y1)/(x2-x1) so f(-3+h) corresponds to y2, f(-3) corresponds to y1, and h corresponds to x2-x1