Question

Help With Calculus

Answer 1

@Shadow @Vocaloid

Answer 2

if cos(x) = 3x/pi somewhere between 0 and pi/2

then

cos(x) - 3x/pi = 0 must have a solution between 0 and pi/2

evaluate cos(x) - 3x/pi at both endpoints (so evaluate it at x = 0 and x = pi/2), and see if 0 is contained between the two results

Answer 3

It is

Answer 4

@SmokeyBrown

Answer 5

then that's it, you're done with the proof

Answer 6

Do I have to make a graph

Answer 7

not really, the algebraic proof should be enough

if your teacher allows it, sure, why not

Answer 8

okay so the evaluation for the pointzero would equal zero and the evaluation at th epoint pi/2 = -1.5

Answer 9

hm close, I think cos(x) - 3x/pi ends up being 1 at x = 0 and -1.5 at pi/2

Answer 10

cos(0) - 3(0)/pi = cos(0) = 1

now, since 0 is between 1 and -1.5, then the intermediate value 0 is a solution, therefore making cos(x) - 3x/pi = 0 somewhere between the interval, therefore making cos(x) = 3x/pi somewhere between the interval, thus completing the proof

Answer 11

if you want to prove this graphically, you can graph cos(x) - 3x/pi between 0 and pi/2 and it will end up crossing the x-axis somewhere, thus showing the proof visually

Answer 12

DO I just make a graph of this polynomial for this question>

Answer 13

you could, yeah, you'd just pick an interval that shows that the function has at least one positive value and one negative value, as well as being continuous

Answer 14

So whats the best way to do this problem like make a graph or another way

Answer 15

I'd probably show it algebraically, you can pick 1 x-value that gives a positive output value, and 1 x-value that gives a negative output value, and, assuming the function is continuous (which it is) that proves that it crosses the x-axis somewhere and thus has at least one root

Answer 16

to save you some time x = 0 and x = 1 fulfill this requirement

Answer 17

so when I substitute 0 I get 1

When I substitute 1 I get -9

Answer 18

good, since one result is + and the other is - that means 0 must be an intermediate value thus proving there's a root between x = 1 and x = 0

Answer 19

Sorry I'm making notes as I go so I'm replying a little late

Answer 20

yeah, np, graphing would work, it's just that making a graph isn't always practical especially on an exam

Answer 21

And btw for the previous question do you just pick any two points and work with it

Answer 22

for the intermediate value problems, yes, I try to pick x-values that are easy to calculate, and work in terms of proving the intermediate value

Answer 23

anyway for 3 you need to take the derivative of sqrt(x+1)

according to derivative rules, bring the exponent down as a coefficient, reduce the exponent by 1, and take the derivative of what's inside the parentheses

Answer 24

SO If I were to pick 0 and 2 It would still work

Answer 25

yes

Answer 26

OKay perfect

Answer 27

I'm confused, what is the exponent in sqrt (x+1)

Answer 28

sqrt(x+1) is equivalent to (x+1)^(1/2) so the exponent would be 1/2

Answer 29

brief review

Answer 30

-1/2?

Answer 31

:S

Answer 32

yeah you're on the right track

so, if we take the exponent down from (x+1)^(1/2)

we get

(1/2)

then we take (x+1)^(1/2), reduce the exponent by 1 to get (x+1)^(-1/2)

Answer 33

then we take the derivative of the thing in parentheses, (x+1). the derivative of a constant by itself is 1 (so derivative of x is 1, derivative of p is 1, derivative of triangle is 1, etc.) and the derivative of a constant is 0, so the derivative of x + 1 is 1 + 0 or just 1

Answer 34

then putting the three parts of the derivative together:

(1/2) * (x+1)^(-1/2) * (1) = your derivative

Answer 35

now it doesn't just want the derivative, it wants f'(1) which is the derivative at x = 1, so we'd just plug x = 1 into the derivative (1/2) * (x+1)^(-1/2) * (1) to get the solution

Answer 36

1/2^3/2

Answer 37

be careful with parentheses

(1/2) * (x+1)^(-1/2) ---> plugging in x = 1 gives us ---> (1/2) * 2 ^(-1/2) = 1/[2sqrt(2)]

Answer 38

basically, a^(-1/2) is equivalent to 1/sqrt(a) since negative exponents can be re-written as positive exponents in the denominator

Answer 39

got it

Answer 40

so the answer to f ' = 1/[2sqrt(2)]

Answer 41

yeah

Answer 42

to show that a piecewise function is differentiable at a certain point, show that that the function is continuous ---> show that the pieces of the function are equal at that point

Answer 43

do I use the x values provided so 0?

Answer 44

yes so in this case just show that

x * sin(1/x) = 0
at x = 0 (which is pretty easy)

Answer 45

its false when I do x=0

Answer 46

well, I guess if you get technical about it, sin(1/x) * x is undefined at 0, but the limit is 0 (it approaches 0 without actually getting there) making the function continuous

Answer 47

i'd have to think about b) a bit more but I think it has to do with the LH and RH limits

Answer 48

Yeha I think it has to do with that too

Answer 49

ugh, idk how I would do it algebraically, but I guess you could graph it

Answer 50

would I graph x sin (1/x)

Answer 51

would I graph x sin (1/x)

Answer 52

@@jhonyy9

Answer 53

you wan graphing xsin(1/x) ?

Answer 54

No, I'm asking if thats What I'm supposed to do

Answer 55

\(\color{#0cbb34}{\text{Originally Posted by}}\) @zarkam21

\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 56

This is my problem, I'm on part b

Answer 57

Q. 4 - ?

Answer 58

Yes Part B

Answer 59

@dude

Answer 60

@zarkam21 than yu see the f(x) for x=0 given that f(x) =0 => not is differentiable

or not ?

Answer 61

Yes

Answer 62

so this mean that for x=0 => f(x) =0 and you need getting the derivativ of zero - so how ?

Answer 63

impossibile - i think is clearly

Answer 64

Okay so how would I use the definition to show that

Answer 65

easy

bc. for x=0 => f(x)=0 => have not derivativ => is not differentiable

i think in this way

Answer 66

OKay thanks, do you care if I get a second opinion?

Answer 67

@Vocaloid when you get a chance can you please look at this

Answer 68

I mean for this the question tells you that x is undefined when \(x sin(\frac{1}{x})=0\)

You cannot find the derivative of an undefined point ¯\_(ツ)_/¯

Answer 69

OKay we can just see what @Vocaloid says, as another opinion because i'm confused myself

So @Vocaloid when you get a change

Answer 70

@Shadow

Answer 71

Undefined then?

Answer 72

x is defined at 0, the function manually defines it to be 0 at x = 0

Answer 73

SO using the definition of a derivative it would be the same

Answer 74

what

Answer 75

The answer to 4b. USing the definition of the derivative, for x=0 => f(x)=0 => not having a derivative. This is not differentiable

Answer 76

that's redundant. "it has no derivative so it's not differentiable"

xsin(1/x) oscillates at x = 0 so it is not differentiable