# Thread: Help me with providing example

1. ## Help me with providing example

Provide an example of function f:R --> R and g:R--> R, and c is subset to R, such that g is not differentiable at c and fog is defferentiable at c.

Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
such that f is not differentiable at g(c) and fog is defferentiable at c.

Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
such that f+g is differentiable at c and f is not defferentiable at c.

it kind of makes me confused

2. Originally Posted by 450081592
Provide an example of function f:R --> R and g:R--> R, and c is subset to R, such that g is not differentiable at c and fog is defferentiable at c.

Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
such that f is not differentiable at g(c) and fog is defferentiable at c.

Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
such that f+g is differentiable at c and f is not defferentiable at c.

it kind of makes me confused
you do mean c is an element of R, right? As in $\displaystyle c \in \mathbb{R}$.

For the first, $\displaystyle c = 0$, $\displaystyle f(x) = 0$ and $\displaystyle g(x) = |x|$

For the second, $\displaystyle c = 0$, $\displaystyle f(x) = |x|$ and $\displaystyle g(x) = 0$

Now try the third.

3. um There is a contracdiction in the second one, they are both saying |0| is differentiable and not defferentible

4. actually, the second example works too, but it seems like you changing the rang of the function though, cause f and g is not an element of R anymore, it becomes {0}

so change the constant function f(x) =0 to f(x) = x

5. Originally Posted by 450081592
um There is a contracdiction in the second one, they are both saying |0| is differentiable and not differentible
no, that is not what the example says.

Originally Posted by 450081592
actually, the second example works too, but it seems like you changing the rang of the function though, cause f and g is not an element of R anymore, it becomes {0}

so change the constant function f(x) =0 to f(x) = x
we are giving two separate examples, it does not matter if the ranges differ in one example from the other.

f and g are real functions. and 0 is an element of $\displaystyle \mathbb{R}$.

f(x) = x does not work in the first example. neither would it work for the second example...

6. Originally Posted by Jhevon
no, that is not what the example says.

we are giving two separate examples, it does not matter if the ranges differ in one example from the other.

f and g are real functions. and 0 is an element of $\displaystyle \mathbb{R}$.

f(x) = x does not work in the first example. neither would it work for the second example...

oh I meant f(x) = X^2, this will work

7. Originally Posted by 450081592
oh I meant f(x) = X^2, this will work
yes. that will work in the first example