Results 1 to 8 of 8

Math Help - functions

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    2

    functions

    Using the graph determine the following
    1)intervals on which the function is increasing
    2) intervals on which the function is decreasing
    3) intervals on which the function is constant



    any help in explaining the steps necessary in solving this question would be appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by zeratul View Post
    Using the graph determine the following
    1)intervals on which the function is increasing
    2) intervals on which the function is decreasing
    3) intervals on which the function is constant



    any help in explaining the steps necessary in solving this question would be appreciated
    A function is said to be increasing over an interval if its first derivative is positive over that interval. i.e. the gradient of the tangents to the curve are positive, i.e sloping upwards.

    A function is said to be decreasing over an interval if its first derivative is negative over that interval. i.e. the gradient of the tangents to the curve are negative, i.e. sloping downwards.

    A function is said to be constant over an interval if its first derivative is 0 over that interval. i.e. the gradient of the tangents to the curve are 0, i.e. parallel to the horizontal.

    Do these definitions help you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    38
    Equation of the shown curve is

    y+4=(x+1)^2

    differentiating both sides wrt. x

    dy/dx=2(x+1)

    for increasing fn. dy/dx>0 i.e. slope of tangent is positive

    2(x+1)>0

    i.e. x>-1

    for decreasing fn. dy/dx<0 i.e. slope of tangent is negative

    2(x+1)<0

    i.e. x<-1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by sumit2009 View Post
    Equation of the shown curve is

    y+4=(x+1)^2

    differentiating both sides wrt. x

    dy/dx=2(x+1)

    for increasing fn. dy/dx>0 i.e. slope of tangent is positive

    2(x+1)>0

    i.e. x>-1

    for decreasing fn. dy/dx<0 i.e. slope of tangent is negative

    2(x+1)<0

    i.e. x<-1
    This assumes that the curve goes off to infinity in each direction. The part of the graph we can see stops at x values of -4 and 2. We can hence, only conclude that the interval for decease is [-4,-1), increase (-1,2], and constant [-1,-1]. These are verifiable by the eye alone without the aid of equations. To assume that the curve is continuous after these intervals is far too presumptuous for mathematical rigor.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    38
    Quote Originally Posted by Mush
    This assumes that the curve goes off to infinity in each direction. The part of the graph we can see stops at x values of -4 and 2. [snip]
    It DOESN'T stops at x=-4 and x=2
    we can see from the graphs that there are ARROW HEADS on both ends which means that it extends from -∞ to +∞
    Last edited by mr fantastic; January 18th 2009 at 01:53 AM. Reason: Added [snip] and quote tags
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by sumit2009 View Post
    This assumes that the curve goes off to infinity in each direction. The part of the graph we can see stops at x values of -4 and 2.

    It DOESN'T stops at x=-4 and x=2
    we can see from the graphs that there are ARROW HEADS on both ends which means that it extends from -∞ to +∞
    Ahh. My eyesight wasn't sharp enough to catch that. Jolly good.

    In any instance, equations aren't necessary. The values can be deduced by graph inspection.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2009
    Posts
    38
    Quote Originally Posted by Mush
    [snip]In any instance, equations aren't necessary. The values can be deduced by graph inspection.

    yes sir I agree with u
    Last edited by mr fantastic; January 18th 2009 at 01:52 AM. Reason: Added [snip] and quote tags.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2009
    Posts
    2
    thanks alot for the replies, but differentiation was not thought as yet so can't use the method given my sumit2009.

    As for the definitions given by Mush thanks alot for them but i'm still not sure how to deduce the answers for the graph, can anyone simplify it a little more for me

    thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 15th 2010, 05:50 PM
  2. Replies: 3
    Last Post: February 23rd 2010, 04:54 PM
  3. Replies: 11
    Last Post: November 15th 2009, 11:22 AM
  4. Replies: 7
    Last Post: August 12th 2009, 04:41 PM
  5. Replies: 1
    Last Post: April 15th 2008, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum