1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Problem with this function continuity and differentiation

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that a and b are real numbers. Find all values of a and b (if any) such that the functions f and g, given by

    a) f(x)={ax+b if x<0 and sin(x) if x≥0}

    b) g(x)={ax+b if x<0 and e2x if x≥0}

    are (i) continuous at 0 and (ii) differentiable at 0

    2. Relevant equations

    3. The attempt at a solution

    So for a) f(x)={ax+b if x<0 and sin(x) if x≥0}

    as the lim→0+

    then sin(0) =0

    as the lim →0-

    then a(0)+b=b

    so hence only when b=0 is the function continuous at 0.

    for a) (ii)

    do I just use the limit definition


    lim h→0+ = limh→0-

    so lim h→0+ (f sin(x+h)-f(sinx)) / h = lim h→0- (f(a(x+h)+b)-f(ax+b))/ h

    lim h→0+ sin h / h= sin 0/0 =0 and lim h→0- = a.

    so it is only differentiable at 0 when a=0

    But the answer says when a=1 and b=0 for a) (ii)

    Please help!
  2. jcsd
  3. Apr 28, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    The derivative of sin(x) at x=0 isn't 0. limit h->0 sin(h)/h isn't 0. sin(0)/0 isn't 0 either. It's undefined.
  4. Apr 28, 2012 #3
    Am I using the wrong approach then?

    What happens if these terms are undefined, does the formula not work?
  5. Apr 28, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper

    limit h->0 sin(h)/h is defined. It's just not 0. You don't have any idea what it might be? Try experimenting with a calculator. You really should know what it is. Hopefully the experiment will remind you.
  6. Apr 28, 2012 #5
    To make f(x) differentiable at x=0 (or in other words, so that the graph doesn't have a corner), ax+b must have the same slope as the derivative of sinx has at x=0, which would be...?
  7. Apr 28, 2012 #6
    ohh right its 1, silly me.
  8. Apr 28, 2012 #7
    Ok so a=1 and b=0.

    But for part b) (ii)

    I have the limit from 0- =a from part a)
    and the lim h-->0+

    = f(e2(x+h))-f(e2x) / h

    = e2x +e2h-e2x / h

    = 1/0 ??

    but the answer says a=2 and b=-1??

    what am I doing wrong now?
  9. Apr 28, 2012 #8


    User Avatar
    Science Advisor
    Homework Helper

    Do you have to everything with difference quotients? You aren't very good at it. If f(x)=e^(2x), f(0)=1 and f'(0)=2. Can't you do it by just differentiating using the chain rule?
    Last edited: Apr 28, 2012
  10. Apr 29, 2012 #9
    yeah I guess so
    But what am I differentiating exactly??

    because Im confused with b=-1

    f(0) for the other function =+b
    f'(0) = a

    DO you have to match these up, because I would say that a=2 and b=1, not -1?
    Im confused?
  11. Apr 29, 2012 #10


    User Avatar
    Science Advisor
    Homework Helper

    I would say a=2 and b=1 as well. There may be an error in your answer key.
  12. Apr 29, 2012 #11
    ok thanks dick!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook