Question

Find the average rate of change for the following functions on the indicated intervals.
f(x)= xcubed - 2x; (0,4)
just don't understand how to work the problem.Help?

Answer 1

HINT: rate of change is it's derivative.

Answer 2

The average rate of change from 0 to 4 is given by


( f(4) - f(0) )/( 4 - 0 )

Answer 3

still dont understand

Answer 4

saifoo.khan, that's the instantaneous rate of change

Answer 5

First find f(4), do you know how to do that?

Answer 6

don't remember

Answer 7

Ah, then what's this called? @jim_thompson5910
What's the main difference? Thanks for correcting.

Answer 8

f(x)= x^3 - 2x


f(4)= (4)^3 - 2*(4) <<--- Evaluate this to find f(4)

Answer 9

but 4 is the y

Answer 10

No, 4 is the x

Answer 11

f(4) means find f(x) when x = 4

Answer 12

For average rate of change... its the gradient of the secant across the interval x = 0 to x = 4

start by finding f(0) and f(4)

then the average rate of change is

\[\frac{f(4) -f(0)}{4 - 0}\]

Answer 13

okay so find the function of 4 and 0?

Answer 14

yes

Answer 15

So this is basically the average? @jim_thompson5910

Answer 16

what now?

Answer 17

So what did you get for f(4)?

Answer 18

40

Answer 19

Try it again, f(4) is NOT 40

Answer 20

56

Answer 21

it's the the average rate of change as stated at the top of the problem saifoo.khan

Answer 22

good, you got it

Answer 23

so f(4) = 56

what about f(0)?

Answer 24

thats 0

Answer 25

right again

Answer 26

so

( f(4) - f(0) )/( 4 - 0 )

then becomes

( 56 - 0 )/( 4 - 0 )

Answer 27

evaluate/simplify that to get your final answer

Answer 28

so 14?

Answer 29

you nailed it

Answer 30

thats the average rate of change?

Answer 31

yes

Answer 32

think of it as the slope of the line that connects the two points (0,0) and (4,56)

Answer 33

you good at word problems?

Answer 34

sure, what's the question?

Answer 35

same directions.
a car travels 360 miles in a period of 180 minutes. find the average velocity of the car in miles per hour over this time period. isn't there a formula for this?

Answer 36

first convert 180 minutes to hours

Answer 37

Use the idea that 1 hour = 60 minutes

Answer 38

3 hours

Answer 39

So the car travels 360 miles in 3 hours


The average velocity is then

(Distance)/(Time)

Answer 40

so 120?

Answer 41

yes, the car's average velocity is then 120 mi/hr

Answer 42

thanks so much! i have lots more to do..are you busy?

Answer 43

no not really, which ones are bugging you the most?

Answer 44

natural logarithms..just don't remember them

Answer 45

A natural log is simply a logarithm, but it has a special base known as 'e'

Answer 46

The value of 'e' is roughly e = 2.718...

Answer 47

5/e^x +1 =1 solve for x

Answer 48

thats whats bugging me..i can't simplify it far enough

Answer 49

\[\Large \frac{5}{e^x}+1=1\] or \[\Large \frac{5}{e^x+1}=1\] ?

Answer 50

yes

Answer 51

which one?

Answer 52

the second

Answer 53

Start by multiplying both sides by e^x+1 to get


\[\Large 5=1(e^x+1)\]

\[\Large 5=e^x+1\]


Is this one of your steps?

Answer 54

yes

Answer 55

What's next?

Answer 56

subtract 1

Answer 57

good

Answer 58

to get what?

Answer 59

but then im stuck

Answer 60

i have 4= e^x

Answer 61

ok great so far

Answer 62

but now what?

Answer 63

now take the natural log of both sides

Answer 64

1.39

Answer 65

e cancels right?

Answer 66

you got it

Answer 67

if the book want's an exact answer, then simply leave it as

x = ln(4)

Answer 68

but if it wants an approximate answer, then you would write

x = 1.39

Answer 69

okay..well..it's a summer assignment..i should be okay just as long as i show work

Answer 70

i gotcha

Answer 71

\[e ^{2x}+2+e ^{-2x}\]

Answer 72

help?

Answer 73

simplifying this? or solving?

Answer 74

factoring

Answer 75

into binomials

Answer 76

hmm one sec

Answer 77

ah ok I think I see what to do

Answer 78

\[e^{2x}+2+e^{-2x}\]

is the same as

\[e ^{2x}+2+\frac{1}{e^{2x}}\]

Answer 79

yeah..just an inverse

Answer 80

exactly

Answer 81

so they cancel?

Answer 82

or make 1

Answer 83

Not quite

Answer 84

common denomenator?

Answer 85

Yes, you need to find the LCD

Answer 86

then get each term to have that LCD, then combine the fractions

Answer 87

so it would be e to the 2x underneath and one fraction would have e^2x squared on top

Answer 88

yes, that's the first fraction

Answer 89

okay..now what

Answer 90

so

\[e ^{2x}+2+\frac{1}{e^{2x}}\]

becomes

\[\frac{\left(e^{2x}\right)^2}{e^{2x}}+2+\frac{1}{e^{2x}}\]

Answer 91

Do the same with the middle term 2

Answer 92

put it over e?

Answer 93

well multiply it by \[\Large \frac{e^{2x}}{e^{2x}}\]

Answer 94

okay.. now what

Answer 95

\[\frac{\left(e^{2x}\right)^2}{e^{2x}}+2+\frac{1}{e^{2x}}\]

then becomes

\[\frac{\left(e^{2x}\right)^2}{e^{2x}}+\frac{2e^{2x}}{e^{2x}}+\frac{1}{e^{2x}}\]

Answer 96

Now combine the fractions

Answer 97

okay

Answer 98

combined