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Math Help - Sketching the graph

  1. #1
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    Sketching the graph

    The following are procedures to draw a graph.

    f'(x)>0 if |x|<2, f'(x)<0 if |x|>2, f'(2)=0, limit as x approaches infinity=1, f(-x)=-f(x), f"(x)<0 if 0<x<3, f"(x)>0 if x>3.

    if the nature of the this function following |x|, what confuses me is how can i draw a graph that is concave downward between 0 and 3, and concave upward greater than 3 when the lines have to be straight.

    -I did try drawing it but i am not sure if it's right. I begin at the origin and draw a line concave downward up to 2, from here should the turning point be pointy like on an |x| graph or smooth like on a parabola? then it continues to be concave downward with a negative gradient until it gets to 3 then it becomes concave upward.

    -Also, for the limit as x approaches infiinity=1, does that mean there is a horizontal asymptote at y=1.
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  2. #2
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    Re: Sketching the graph

    Hey johnsy123.

    An example of a sudden change from concave upward to downward is that of a circle (although it technically isn't a function), but if you look at when y = 0, this is an example of something going from concave up to concave down.

    An example of a function doing this would be if you the two semi-circles (one below the y axis and one above) and you shifted the one below the y-axis so that the left-hand point on the bottom semi-circle matched up with the right-hand side point of the top-circle (kind of like a skate ramp kinda thing): this is an example of a function going from concave down to concave up.

    As for the horizontal asymptote, I think this is the case but the asymptote is going to be bounded from above and not below since the second derivative is always positive from x > 3 onwards.
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  3. #3
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    Re: Sketching the graph

    I am aware of the nature of concavity. What i am buzzed on is should i sketch this graph like a |x| graph where the lines are linear? but if i do that then wouldn't it be impossible to draw concave downward and upward?

    does the "f'(x)>0 if |x|<2" mean that because the |x| is there i have to draw this graph like that of a |x|?
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    Re: Sketching the graph

    here's what i tried anyway

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  5. #5
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    Re: Sketching the graph

    Quote Originally Posted by johnsy123 View Post
    The following are procedures to draw a graph.

    f'(x)>0 if |x|<2, f'(x)<0 if |x|>2, f'(2)=0, limit as x approaches infinity=1, f(-x)=-f(x), f"(x)<0 if 0<x<3, f"(x)>0 if x>3.

    if the nature of the this function following |x|, what confuses me is how can i draw a graph that is concave downward between 0 and 3, and concave upward greater than 3 when the lines have to be straight.

    -I did try drawing it but i am not sure if it's right. I begin at the origin and draw a line concave downward up to 2, from here should the turning point be pointy like on an |x| graph or smooth like on a parabola? then it continues to be concave downward with a negative gradient until it gets to 3 then it becomes concave upward.

    -Also, for the limit as x approaches infiinity=1, does that mean there is a horizontal asymptote at y=1.
    What it's saying is, that your function has the property \displaystyle \begin{align*} f'(x) > 0 \end{align*} for all \displaystyle \begin{align*} |x| < 2 \end{align*}, in other words, it has the property \displaystyle \begin{align*} f'(x) > 0 \end{align*} when \displaystyle \begin{align*} -2 < x < 2 \end{align*}.

    It also has the property that \displaystyle \begin{align*} f'(x) < 0 \end{align*} if \displaystyle \begin{align*} |x| > 2 \end{align*}, in other wrods, that it has the property \displaystyle \begin{align*} f'(x) < 0 \end{align*} when \displaystyle \begin{align*} x < -2 \end{align*} and when \displaystyle \begin{align*} x > 2 \end{align*}.

    Clearly that means \displaystyle \begin{align*} f'(x) = 0 \end{align*} when \displaystyle \begin{align*} |x| = 2 \end{align*}, in other words, has stationary points where \displaystyle \begin{align*} x = \pm 2 \end{align*}.


    Another important property is that \displaystyle \begin{align*} f(x) \to 1 \end{align*} as \displaystyle \begin{align*} x \to \infty \end{align*}, so has a horizontal asymptote at \displaystyle \begin{align*} y = 1 \end{align*}.


    It's an odd function, because \displaystyle \begin{align*} f(-x) = -f(x) \end{align*}, that means there is a reflection in BOTH the x and y axes.


    Finally you're told that \displaystyle \begin{align*} f''(x) < 0 \end{align*} if \displaystyle \begin{align*} 0 < x < 3 \end{align*} and \displaystyle \begin{align*} f''(x) > 0 \end{align*} if \displaystyle \begin{align*} x > 3 \end{align*}. Recall that if the second derivative is negative at a stationary point, the stationary point is a local maximum. So that means at the point where \displaystyle \begin{align*} x = 2 \end{align*} you have a local maximum.


    From the property of being an odd function, what does that tell you about the stationary point at \displaystyle \begin{align*} x = -2 \end{align*}?
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  6. #6
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    Re: Sketching the graph

    The stationary point at x=-2 is just the same as at x=2?
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