Question

Calculus. Limit Question

Answer 1

I have a piece wise function

Answer 2

2x^3 - 3, x is less than and equal to 1
6x-4, x is greater than 1

Answer 3

Which of the following statements is true?
I. Limit as x approaches 1 f(x) exists
II. Limit as x approaches 1 f ' (x) exists
III. F'(1) exists. Sorry that should be a lower case f

Answer 4

Statement one is false since when I sub (1)
Into both I get 2 difference values -1 and 2

Answer 5

When I take the derivative of each one and I sub 1 I get the same number which is 6

Answer 6

Nice, and what is the problem ?

Answer 7

Not sure about third statement

Answer 8

If there is no point at x=1, how can there be an insta nteneous slope at x=1?

Answer 9

Cool. Ok so statement 2 is the only true one, correct?

Answer 10

Yes.
For something to be differentiable, it has to be continuous. Here, you noted that this function isn't really continuous.

Answer 11

yes, so while your f'(x) approaches the same value as x=1, your derivative exactly at x=1 is not defined (simply because there is no point at x=1, and thus you don't have an instanteneous slope (the derivative) at x=1).

Answer 12

clear on that ?

Answer 13

No point at \(x=1\)? What does that mean? \(f(1) =-1\).

Answer 14

Guess I got confused because I am testing out a new graphing calculator and the piece wise function looks continuous.

Answer 15

Parth,

\(\color{#000000}{ \displaystyle 6(1)-4=6-4=2 }\)
\(\color{#000000}{ \displaystyle 2(1)^3-2=2-3=-1 }\)

Answer 16

Also, got confused because of some theorem that states if a function is differentiable then it is continuous

Answer 17

But I do have to look at all of the pieces of a piece wise function to make sure they connect at the same location

Answer 18

Yes, that is exactly true, precal. We can explain the theorem if you lie ...

Answer 19

if you like, (not lie, sorry)

Answer 20

Sure

Answer 21

The function is defined as \(-1\) at \(x=1\). I don't understand by "no point at x=1".

Answer 22

Also, in the inequality the equal sign is on the first equation
2x^3 - 3. Note x equals 1 and x is less than 1

Answer 23

PatthKohle, you have a piece wise function that consists of two peaces.
2x^3 - 3, x≤1
6x-4, x>1

for the first peace
f(1)=2(1)^2-3=2-3=-1

for the second peace
f(1)=6(1)-4=2

Answer 24

Now, as far as that theorem is concerned ...

Answer 25

why are you using the second piece for x = 1? that is only applicable to x > 1

Answer 26

@ParthKohli because if those numbers aren't the same that means there is a break in the graph

Answer 27

yes, that's what discontinuity is. but that doesn't mean that the function isn't defined at x=1.

Answer 28

Would the limit of f(x) as x-->1 exist? No, not a 2-sided limit.
Would the limit of f'(x) as x-->1 exist? yes, we all agreed on that.
Would f'(x) at x=1 exist? Tricky, but actually not.

Answer 29

You might have noticed when learning (for example) the mean vaue theorem that it demand that the function be continunous on [a,b] and differntiable on (a,b)....

Why? bec ause end-points don't have instaneneous slope....

Answer 30

so finishinf off with III,
your derivative f'(1) (if f(1)=f(1) from both sides) won't exist, since that endpoint will not have an instanteneous slope.

Answer 31

If \(\color{#000000}{ \displaystyle f(x) }\) is differentiable, then (by definition of "differentiable") an instanteneous slope on every point on \(\color{#000000}{ \displaystyle f(x) }\) exists.

If each point has an instaneneous slope, then without a total continuity, there would be at least one point when instanteneous slope does not exist.