Find f'(0), if it exists and is f continuous at x=0

[MarcL]

*couldn't edit the title so Find*

Homework Statement

f(x) = x+1 , x<0
f(x) = 1 , x=0
f(x)= x²-2x+1 , x>0

The Attempt at a Solution

for a (find f'(0), if it exists) i did as followed

Lim h->0^- (x+h)+1-(x) /h giving me in the end 1

as for the third equation, I did as follow:
Lim h->0⁺ (x+h)²-2(x+h)+1 - (x²-2x+1) /h

in the end giving me 0
So I answered that f'(0) doesn't exsit

for b) is f continuous at x=0
i just solved the limit of x +1 which gave me 1 and the limit of x²-2x+1 which gave me 1. Hence I answered that f is continuous at x=0

If I need to add more information, let me know =)

 

[SammyS]

*couldn't edit the title so Find*

Homework Statement

f(x) = x+1 , x<0
f(x) = 1 , x=0
f(x)= x²-2x+1 , x>0

The Attempt at a Solution

for a (find f'(0), if it exists) i did as followed

Lim h->0^- ((x+h)+1-(x)) /h giving me in the end 1

as for the third equation, I did as follow:
Lim h->0⁺ ((x+h)²-2(x+h)+1 - (x²-2x+1)) /h

in the end giving me 0
So I answered that f'(0) doesn't exsit

for b) is f continuous at x=0
i just solved the limit of x +1 which gave me 1 and the limit of x²-2x+1 which gave me 1. Hence I answered that f is continuous at x=0

If I need to add more information, let me know =)

That all looks good. --- except I inserted some parentheses, where needed.

 

[HallsofIvy]

By the way, for general functions the fact that $\lim_{x\to a^-}f(x)\ne \lim_{x\to a^+}f(x)$ does not prove that f(a) does not exist- only that f is not continuous at x= a. However, the derivative of a function, while not necessarily continuous, does have the "intermediate value property". That is, if f'(a)= A and f'(b)= B then, for any C between A and B, there exist c between a and b such that f(c)= C.

That implies that, whether f' is continuous or not, if f'(a) exists, we must have $\lim_{x\to a^-}f(x)= \lim_{x\to a^+}f(x)$ so your argument is valid.

 

[MarcL]

I don't understand why you brought the "intermediate value" theory in this problem. Isn't used to find if f(x) as a root between a and b considering n isn't equal to a or b?

 

[vela]

*couldn't edit the title so Find*

Homework Statement

f(x) = x+1 , x<0
f(x) = 1 , x=0
f(x)= x²-2x+1 , x>0

The Attempt at a Solution

for a (find f'(0), if it exists) i did as followed

Lim h->0^- (x+h)+1-(x) /h giving me in the end 1

as for the third equation, I did as follow:
Lim h->0⁺ (x+h)²-2(x+h)+1 - (x²-2x+1) /h

in the end giving me 0
So I answered that f'(0) doesn't exsit

You evaluated the second limit wrong somehow. It should equal -2, not 0.

for b) is f continuous at x=0
i just solved the limit of x+1 which gave me 1 and the limit of x²-2x+1 which gave me 1. Hence I answered that f is continuous at x=0

If I need to add more information, let me know =)

You need to show
$$\lim_{x \to 0} f(x) = f(0)$$What you've written shows that
$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)$$which means
$$\lim_{x \to 0} f(x)$$exists. You should probably state explicitly that the limit equals f(0) so the grader knows you understand what continuity means.

 

[SammyS]

You evaluated the second limit wrong somehow. It should equal -2, not 0.
...

How did I miss that ? DUH !

 

[MarcL]

Lim _h->0⁺ ((x+h)²-2(x+h)+1 - (x²-2x+1)) /
is then x²+2xh+h² -2x+2h+ 1 - x²+2x-1
which can then be simplified to 2xh+h^2+2h
so Lim _h->0⁺ (2xh+h²+2h)/h
And here its just basic simplifying however.... I think its where I usually mess up (small mistakes...) and I did Lim _h-->0⁺ 2xh + h + 2h giving in the end 2x(0) + (0) + 2 (0)

 

[vela]

Lim _h->0⁺ ((x+h)²-2(x+h)+1 - (x²-2x+1)) /
is then x²+2xh+h² -2x-2h+ 1 - x²+2x-1

You flipped a sign.

which can then be simplified to 2xh+h^2+2h
so Lim _h->0⁺ (2xh+h²+2h)/h
And here its just basic simplifying however.... I think its where I usually mess up (small mistakes...) and I did Lim _h-->0⁺ 2xh + h + 2h giving in the end 2x(0) + (0) + 2 (0)

 

[HallsofIvy]

I don't understand why you brought the "intermediate value" theory in this problem. Isn't used to find if f(x) as a root between a and b considering n isn't equal to a or b?

Mathematical theorems can be used for more than one purpose! Ignoring the reference to the "intermediate value theorem" do you understand my point that finding $\lim_{x\to a^+}f(x)$ and $\lim_{x\to a^-} f(x)$ and finding that they are the same does not necessarily tell you what f(a) is?