Question

This question is from a James Stewart Calculus book: Suppose f is a function that satisfies the equation f(x+y)=f(x)+f(y)+x^2y+xy^2 for all real numbers x,y. Also suppose as x->0 , f(x)/x->1. Then we have 3 questions: a,b, and c.
a. Find f(0).
b. Find f'(0).
c. Find f'(x).

Answer 1

nice question, hmmm

Answer 2

ooh ooh i got the first one!

Answer 3

oops let me trying posting that one more time

Answer 4

i got \(f(0)=0\) right?

Answer 5

right

Answer 6

(d) deals with partial derivatives?

Answer 7

Never mind we can use l'hopitals rule

Answer 8

This is all the problem gave.
But I can tell you partial derivatives have not been covered before where this question is placed in the book.

Answer 9

I'm actually having trouble determine if f'(0) exists or not.
Like I found a contradiction in something I was doing.

Answer 10

I know f'(0)=1 or f'(0) dne.

Answer 11

I assumed f'(0)=1
but then I got a contradiction which led me to believe f'(0) dne

Answer 12

But I can't post my latex because I can't find a problem in what I typed yet

Answer 13

\[ \text{ We have } f(x+y)=f(x)+f(y)+x^2y+xy^2 \\
\text{ Let } y=0 \text{ so } f(x)=f(x)+f(0) \text{ => } f(0)=0 \\
\text{ Since } \lim_{x \rightarrow 0} \frac{f(x)}{x}=1 \]

\[ \text{ and we have} f(0)=0 \text{ then we also have } \\
\lim_{x \rightarrow 0 } f’(x)=1 \\
\text{ From this, I can only tell either we have } f’(0)=1 \text{ or } \\ f’(0)
\text{ does not exist} \\
\]

\[ \text{ I also know } f \text{ is an odd function: } \\
\text{ let } y=-x \\
f(x+(-x))=f(x)+f(-x)+x^2(-x)+x(-x)^2 \\
f(0)=f(x)+f(-x) \\
0=f(x)+f(-x)\\
-f(x)=f(-x)\\

\]

\[\text{ But I don’t think this helps me to determine f’(0) } \\
\text{ I think } f
\text{ is continuous because it said that equation was for all real numbers}\\
u=x+y \\
\text{ but I’m not sure } f \text{ is differentiable for all real numbers } u \\
\text{ We also know from } f \text{ being odd that } \\
f’ \text{ will be even } \\
\]

\[ \text{ So I’m going to differentiate } f \text{ now } \\
f’(x+y)=f’(x)+f’(y)+x^2+y^2+4xy \\
\text{ Let } y=-x \text{ so we have assuming that } f’\text{ exists at } 0 \\
f’(x+(-x))= f’(x)+f’(-x)+x^2+x^2-4x^2=2f’(x)-2x^2 \\
\text{ so we have } f’(0)=2f’(x)-2x^2 \\
\text{ so } f’(x)=\frac{ f’(0)+2x^2 }{2}\\
\]

\[ \text{ which means going back to } \\
\lim_{x \rightarrow 0 } f’(x)=1 \\
\text{ we have } \\
\lim_{x \rightarrow 0 } \frac{ f’(0)+2x^2 }{2}=1 \\
\lim_{x \rightarrow 0 } (f’(0)+2x^2)=2 \\
\text{ but if } f’(0)=1 \text{ then we have a contradiction because} \\
1+0 \neq 2 \\
\text{ So this leads me to think } f’(0) \text{ does not exist } \]

\]

Answer 14

I did not say derivative of y is y'
because I think it doesn't depend on any number
like I'm saying y is not a function of x

Answer 15

or I mean on any variable (number ) whatever

Answer 16

wait ...
f'(0) doesn't have to be 1 or does not exist
it could be some other number

Answer 17

if I wanted f'(0) to exist it would have to be 2

Answer 18

in order for that one limit to hold

Answer 19

\[\lim_{x \rightarrow 0}(f'(0)+2x^2) =2 \]
since x->0 then 2x^2->0
so f'(0) is 2..
so f'(x)=1+x^2

Answer 20

Let me know if you guys disagree on anything

Answer 21

im still at half way of the work, i like the odd function part xD

Answer 22

i do too

Answer 23

I thought it was the cutest thing about the problem

Answer 24

lol but if f'(x)=1+x^2
then f'(0)=1+0=1 not 2
so f'(0) does not exist?

Answer 25

this is really good

Answer 26

well my answer is not really good

Answer 27

but i think the problem is good

Answer 28

The weird thing is if you have completed cal 1 all the way up passed implicit differentiation you should be able to do this problem since it is right after the chapter 3 review.

Answer 29

So far what you did is really nice^_^

Answer 30

I think you guys get buzzed off of math lol

Answer 31

myininaya i am struggling with understanding below derivative -

\[\text{ So I’m going to differentiate } f \text{ now } \\
\color{Red}{f’(x+y)=f’(x)+f’(y)+x^2+y^2+4xy }\\
\text{ Let } y=-x \text{ so we have assuming that } f’\text{ exists at } 0 \\
f’(x+(-x))= f’(x)+f’(-x)+x^2+x^2-4x^2=2f’(x)-2x^2 \\
\text{ so we have } f’(0)=2f’(x)-2x^2 \\
\text{ so } f’(x)=\frac{ f’(0)+2x^2 }{2}\\\]

how do we differentiate when with respect to (x+y) ?

Answer 32

I was feeling icky about that part.

Answer 33

I wasn't sure about it.
I was looking at if as if f depended on x and y
Like f(x)=x^2 we could say this f'(x)=2x
or like if w ehad f(y)=y^2 we could say f'(y)=2y

Answer 34

If that makes sense

Answer 35

But I do know we have f(x+y)
and if f(x+y)=x+y
then f'(x+y)=1

Answer 36

so yeah I think want to scratch that

Answer 37

I thought you have deduced that part using the fact that f'(x) is even, but still trying to look for strong evidence

x^2y + xy^2 = xy(x+y)

we need to use partials or some other means to break it down i think

Answer 38

This question comes way before partials though

Answer 39

ok I think I got it...

Answer 40

maybe

Answer 41

\[f(\frac{x}{2}+\frac{x}{2})=f(\frac{x}{2})+f(\frac{x}{2})+(\frac{x}{2})^2\frac{x}{2}+\frac{x}{2}(\frac{x}{2})^2\]

Answer 42

not we can differentiate with respect to one variable

Answer 43

so we have
\[f'(x)=f'(\frac{x}{2})+\frac{3x^2}{2}\]

Answer 44

Oh nice, but we can't plugin x=0 :/

Answer 45

yeah so sad

Answer 46

i'm thinking maybe we could use the f(x)/x->1 as x->0 more ...
maybe

\[f'(0)=\lim_{x \rightarrow 0}\frac{f(x)-f(0)}{x-0}=1 \]
so f'(0) is 1

Answer 47

since f(0) is 0

Answer 48

so now we need to find f'(x)

Answer 49

Wow! that looks like a clever idea

Answer 50

we can use all our previous discoveries here i feel

Answer 51

something about f being off and f' being even?

Answer 52

\[ \large f'(x)=\lim_{x \rightarrow y}\frac{f(x)-f(y)}{x-y }\]

Answer 53

or maybe let x->-y
so that we can use f(-y) = -f(y) to our advantage :

\[ \large f'(x)=\lim_{x \rightarrow -y}\frac{f(x)-f(-y)}{x-(-y) }\]

\[ \large f'(x)=\lim_{x \rightarrow -y}\frac{f(x)+f(y)}{x+y }\]

Answer 54

\[ \large f'(x)=\lim_{x \rightarrow -y}\frac{f(x+y
) - xy(x+y)}{x+y }\]

\[ \large f'(x)=\lim_{x \rightarrow -y}\frac{f(x+y
)}{x+y } - xy \]

\[ \large f'(x)=1 +y^2\]

Answer 55

how did we endup f'(x) in terms of y hmm

Answer 56

well both of your limits tend to y

Answer 57

I think it might be f'(y)=1+y^2
so f'(x)=1+x^2
which is what I actually had earlier doing things in an illegal way

Answer 58

I see, need to check this with fresh mind in the morning again. thanks for the beautiful question myininaya :)

Answer 59

Thanks for looking at @ganeshie8

Answer 60

I like looking at the problem plus section at the end of each section in the cal 1-3 book