Continuity/differentiation of piecewise functions

G

Gascoigne

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I'm not sure on the methods for solving the parts of this question. Help appreciated.

f(x) = sqrt(x+1) for 0 ? x ? 3
5 – x for 3 < x ? 5

a) Is f continuous at x = 3? Explain why or why not.
Definition of continuity, limit exists and f(c) exists.

b) Find the average rate of change on the closed interval 0 ? x ? 5
Secant line of the first piece averaged with the secant line of the second piece?
Or, the slope of the first secant line averaged with the second?

c)Suppose the function g is defined by
g(x) = k*sqrt(x+1) for 0?x?3
mx + 2 for 3<x?5

where k and m are constants. If g is differentiable at x = 3, what are the values of k and m?
I don't know where to start on this.
 
a) A function is continuous if you can trace it with your finger without removing your finger from the curve.

y=x^1/2 for 0<x<3
square both sides
y^2=x a parabola, open right, vertex at 0,0 for 0<x<3

y=5-x for 3<x<5
a straight line, slope -1, y intercept at 5

at x=3
parabolas y value= sqrt3
lines y value 2

you would have to remove your finger at x=3 to continue tracing the curve.
function is not continuous

rest to follow

Arthur
 
c)
we defined continuous function in a). You can trace it without removing your finger from the curve.
Differentiable function has the same value for the derivative regardless of how you approach the point.

y=k[x-1]1/2 for 0<x<3
y=mx+2 for 3<x<5
if the function is differentiable at x=3 , we will get the same value for the derivative if we approach the point from 2.9999 or from 3.0001

from the left 0<x<3
y=k[x-1]^1/2 take derivative
y' = 1/2 k [x-1]^(-1/2)
evaluate at x=3
y ' = 1/2 k /sqrt2
y ' = k/[2sqrt2]

from the right 3<x<5
y=mx+2 yake derivative
y ' = m

for the function to be differentiable,the y ' at x=3 should be the same.
set y'=y'
k/[2 sqrt 2] =m
k= 2 sqrt2 m answer



b) I do not know how your instructor defined ' average' for me to do b. SORRY

Arthur
 

Gascoigne said:
I'm not sure on the methods for solving the parts of this question. Help appreciated.

f(x) = sqrt(x+1) for 0 ? x ? 3
5 – x for 3 < x ? 5

a) Is f continuous at x = 3? Explain why or why not.
Click to expand...


Hello Gascoigne!

Tracing the graph of this function with your finger provides an intuitive feel for whether or not f is continuous along its entire domain. But, that method requires you to first draw the graph, and it is not the definition of continuity. I suspect that an instructor would like to see you use the definition.


Definition of continuity, limit exists and f(c) exists.
Click to expand...


Be cautious when you write a definition! Your definition is not correct, for two reasons.

1) What is c?

There is no symbol c in part (a), and you did not define it, either.

2) The existence of the limit at x = 3, as well as the existence of f(3), do NOT guarantee continuity of f at x = 3.

The definition of continuity at a point requires equality of a limit and a function value.

If the following EQUATION is true, then f is continuous at x = 3.

\(\displaystyle \lim_{x \to 3} f(x) \;=\; f(3)\)

Note that the truthfulness of this equation implies two things.

1) The limit exists

2) f is defined at x = 3

Note that the truthfulness of this equation explicitly states that the limit and the function value are equal. This is crucial for continuity to exist because there are many different functions where their limit exists, and they exist, yet continuity fails to exist.



Do you know how to calculate the two-sided limit of f as x approaches 3?



If you need help, then please show any work that you're able to accomplish, and try to say something about why you're stuck.

I'll hold off on parts (b) and (c) until you finish part (a).

Cheers,

~ Mark :)



MY EDIT: corrected an inaccurate statement that I made regarding a point versus a value of the independent variable
 
arthur ohlsten said:
a) A function is continuous if you can trace it with your finger without removing your finger from the curve.

y=x^1/2 for 0<x<3 This is not the correct definition of f

square both sides Do you realize that you're now working with f[sup:j9sapz5t]2[/sup:j9sapz5t]? It's better to stick with the original function.
Click to expand...
 
The definition of continuity,[Calculus and Analtic Geometry,Abraham Schwartz pg 395]
"The function f[x] is continuous at x=a if 1) f[a] exists, and if 2) for every e>0, no matter how small, there is a neighberhood of x=a such that I f[x]-f[a]I <e for all x of the neighberhood"

When the definition of limit is kept in mind our definition, students usually understand the definition in terms of, " If you can trace the graph without removing your finger from the graph ,the function is continuous"

I did not realize the student had to "prove " the function was continuous, but only had to determine if it was, and indicate where it was discontinuous. But I have made many errors and SORRY I misunderstood the problemn

SORRY
Arthur
 
mmm4444bot said:

Gascoigne said:
I'm not sure on the methods for solving the parts of this question. Help appreciated.

f(x) = sqrt(x+1) for 0 ? x ? 3
5 – x for 3 < x ? 5

a) Is f continuous at x = 3? Explain why or why not.
Click to expand...


Hello Gascoigne!

Tracing the graph of this function with your finger provides an intuitive feel for whether or not f is continuous along its entire domain. But, that method requires you to first draw the graph, and it is not the definition of continuity. I suspect that an instructor would like to see you use the definition.


Definition of continuity, limit exists and f(c) exists.
Click to expand...


Be cautious when you write a definition! Your definition is not correct, for two reasons.

1) What is c?

There is no symbol c in part (a), and you did not define it, either.

2) The existence of the limit at x = 3, as well as the existence of f(3), do NOT guarantee continuity of f at x = 3.

The definition of continuity at a point requires equality of a limit and a function value.

If the following EQUATION is true, then f is continuous at x = 3.

\(\displaystyle \lim_{x \to 3} f(x) \;=\; f(3)\)

Note that the truthfulness of this equation implies two things.

1) The limit exists

2) f is defined at x = 3

Note that the truthfulness of this equation explicitly states that the limit and the function value are equal. This is crucial for continuity to exist because there are many different functions where their limit exists, and they exist, yet continuity fails to exist.



Do you know how to calculate the two-sided limit of f as x approaches 3?



If you need help, then please show any work that you're able to accomplish, and try to say something about why you're stuck.

I'll hold off on parts (b) and (c) until you finish part (a).

Cheers,

~ Mark :)



MY EDIT: corrected an inaccurate statement that I made regarding a point versus a value of the independent variable
Click to expand...

I was taught the equation with f(c) representing the function value at point c. My book uses the same notation, so I thought it was standard. I do understand how to determine continuity and a limit, I just forgot to type that the function value has to equal the limit value at the given point.


arthur ohlsten said:
Differentiable function has the same value for the derivative regardless of how you approach the point.
Click to expand...

Thanks, I was missing part of the differentiable function definition.


Also, the question is from the college board, so the definition of "average rate of change" is their definition. If you use the difference quotient to find the secant line of the first and second pieces of the function, you have the average rate of change for the two pieces, I think. I don't know what you do after that, though. The average rate of change for the second part is -1. Is the average rate of change of a piecewise function piecewise? i.e., is the average rate of change for this function the rate of change for the equation used for the first interval over that interval, and the rate of change of the equation used for the second interval over that interval? Such that,
rate of change for 0 ? x ? 5 is
1/2(x+1)^-1/2 for 0 ? x ? 3 and
-1 for 3 < x ? 5 ?

A piecewise definition for a piecewise function utilizing two intervals that are subsets of the given interval. It seems logical, but I have no idea if it's right.
 
I did not mean to imply that graqphing the function was neccessary or that tracing the function with your finger was the definition of continuity. I told the student that it was a method of determining continuity.
When a student asks for the derivative of x^n, we usually tell him nx^(n-1) and not a derivation of the derivative.
Students are taught that y^2=x is a parabola, but we do not teach y=sqrtx is a parabola.
I squared both sides so it would be obvious to a student that y=sqrt x is a parabola. In mathematics we are usually trying to reduce a problem to something we know how to handle
If the original was y=I sqrt x I , I would have pointed out this was the upper portion of a parabola.
If the equation was y= - Isqrtx I i would have said bottom half of a parabola
The equation was y=sqrt x , thus a parabola, but I was not sure the stident "saw" that.

Arthur
 
arthur ohlsten said:
I did not mean to imply that graqphing the function was neccessary or that tracing the function with your finger was the definition of continuity. I told the student that it was a method of determining continuity.
When a student asks for the derivative of x^n, we usually tell him nx^(n-1) and not a derivation of the derivative.
Students are taught that y^2=x is a parabola, but we do not teach y=sqrtx is a parabola.
I squared both sides so it would be obvious to a student that y=sqrt x is a parabola. In mathematics we are usually trying to reduce a problem to something we know how to handle
If the original was y=I sqrt x I , I would have pointed out this was the upper portion of a parabola.
If the equation was y= - Isqrtx I i would have said bottom half of a parabola
The equation was y=sqrt x , thus a parabola, but I was not sure the stident "saw" that.

Arthur
Click to expand...
I don't know how to make it any clearer that I do in fact understand continuity. I appreciate the explanation, but I indicated that I knew how to solve that part of the question right from the start (I just accidentally left out part of the definition). The function is in fact continuous at x = 3. You left out a term in your explanation. The function was sqrt(x+1), not sqrt(x). If you plug in 3 the point exists, the limit exists, and they are both equal to 2, satisfying the definition.

Also, for part c) you used a minus sign instead of a plus sign for the original function. The derivative, then, would be 1/2k^1/2. At x = 3 that reduces to 1/2k(1/2). The limit of the derivative from the left and the right must be equal, so 1/2k(1/2) = m. I've found k and m in terms of each other, but I don't know how to solve for a numerical value.
 
I am sorry for the errors I made, but it was probably late and I never checked my work. Thank [God I am no longer a student]
It is obvious you understand the concept of continuity and of the derivitive.
If you have a equation relating m and k, there is no single value for the parameters, but rather an infinite set of values.
You can plot k=am, a straight line, and any point on the line satisfys the requirement for the function.

I am sorry for the errors I made, but I tried to show a method to see if the function was continuous . Students usually find this of value, similiar to high school SOHCAHTOA or BAD BOYS RAPE OUR YOUNG GIRLS BUT VIOLET GIVES WILLINGLY, from my days as a radar technician, or ROY G BEV for colors.

The same for my indicating that you must approach the point from 2.99999 and from 3.000
only to help students visualize what the definition means. Again students usually understand the math easier if they observe this definition.

Sorry if I confused you, and for the length of the explanation. I must admit I never taught undergraduates , only graduate students and that was over 50 years ago.

Arthur
 
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