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Math Help - Graph sketching, continuity, differentiation - Is this right????

  1. #1
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    Graph sketching, continuity, differentiation - Is this right????

    (I would appreciate either an explanation as to how to solve the problems, or the correct answer so I could figure out what I did wrong.)

    1. Let f(x) =
    4-x^2 if x < 2
    x^2 - 5x + 6 if 2 < x < 3
    x if x > 3
    a) Sketch the graph of f.

    http://img293.imageshack.us/my.php?image=calcgraphgk1.png


    b) At what numbers is f continuous?

    - I'm not sure if I did this right, are you supposed to do something with limits to show what numbers are continuous?
    (-∞, 2], [2, 3], [3, ∞)

    c) At what numbers is f differentiable?

    - Again, I've seen in some places where your answer has to include limits. Do you have to use limits in your work or something?
    (-∞ ,∞)

    d) Give a formula for f'.
    f'(x) =
    2x if x < 2
    2x - 5 if 2 < x < 3
    1 if x > 3
    2. Let f(x) = x^3 - 9x^2 + 24x + 3
    a) For what values of x does the graph of f have a horizontal tangent
    - Okay I have no idea what to do for this one. I tried looking through my textbook and couldn't find anything. If someone could show me how to do this, that would be great.

    b)
    f''(x) = 9
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  2. #2
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    graph seems to be right...


    f is continuous for all x except when u see a "jump" discontinuity or it goes to infinity... so you will get -infinity to 3 right and then 3 to positive infinity. not incl 3 itself.

    think about when your function isnt differentiable... think about the definition of a derivative and i think that will lead u to ur next answer.


    Check your first derivative.


    The second problem... Take the derivative of f(x) , set it to zero and solve for x.
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  3. #3
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    so for the 2nd problem, ill give u a lil help
    Let f(x) = x^3 - 9x^2 + 24x + 3

    so

    f'(x) = 3x^2 - 18x + 24

    set it equal to zero

    x^2 - 6x + 8 = 0

    (x-2)(x-4) = 0

    thus x = 2 and 4 gives u a horizontal tangent b/c the first derivative is = 0

    this takes me way back lol
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  4. #4
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    Quote Originally Posted by AntiSandman View Post
    (I would appreciate either an explanation as to how to solve the problems, or the correct answer so I could figure out what I did wrong.)

    1. Let f(x) =
    4-x^2 if x < 2
    x^2 - 5x + 6 if 2 < x < 3
    x if x > 3
    a) Sketch the graph of f.

    http://img293.imageshack.us/my.php?image=calcgraphgk1.png


    b) At what numbers is f continuous?

    - I'm not sure if I did this right, are you supposed to do something with limits to show what numbers are continuous?
    (-∞, 2], [2, 3], [3, ∞)

    c) At what numbers is f differentiable?

    - Again, I've seen in some places where your answer has to include limits. Do you have to use limits in your work or something?
    (-∞ ,∞)

    d) Give a formula for f'.
    f'(x) =
    2x if x < 2
    2x - 5 if 2 < x < 3
    1 if x > 3
    2. Let f(x) = x^3 - 9x^2 + 24x + 3
    a) For what values of x does the graph of f have a horizontal tangent
    - Okay I have no idea what to do for this one. I tried looking through my textbook and couldn't find anything. If someone could show me how to do this, that would be great.

    b)
    f''(x) = 9
    For 1. b) You have a hybrid function made up of three polynomial functions. Polynomials are defined and continuous over \mathbf{R} unless otherwise specified. So your function may only be discontinuous at x=2 and x=3.

    To see if they are continuous, see what value the function approaches from the left and the right of this point.

    So for x=2, the function will approach 0 from the left (as 4-2^2 = 0) and 0 from the right (as 2^2 - 5 \times 2 + 6 = 0. And since the function is defined at [/tex]x=2[/tex], we can say the function is continuous at x=2.

    However, for x=3, from the left the function approaches 0 as 3^2 - 5 \times 3 +6 = 0 but from the right the function approaches 1. Since the left hand limit does not equal the right hand limit, the function is discontinuous at that point.

    So the function is continuous for x \in \mathbf{R} \backslash \{3\}.
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