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Math Help - Sketching graphs

  1. #1
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    Exclamation Sketching graphs

    Hi guys

    I've been doing revision for yearlies and I've been having difficulties figuring out these questions:

    1) Sketch the graph of the periodic function such that f(x) = x for 1 < x ≤ 1 where the period of f(x) is 2.

    2) Find the largest possible domain and range of the function f(t) = abs(t) t.

    3) If f(x) = x^2 6x + 5, sketch a graph of the region defined by the intersection of the inequalities f(x) + f(y) ≤ 0 and f(x) f(y) ≥ 0.

    Could you please help me out?

    Thanx
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  2. #2
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    Hello xwrathbringerx
    Quote Originally Posted by xwrathbringerx View Post
    Hi guys

    I've been doing revision for yearlies and I've been having difficulties figuring out these questions:

    1) Sketch the graph of the periodic function such that f(x) = x for 1 < x ≤ 1 where the period of f(x) is 2.

    2) Find the largest possible domain and range of the function f(t) = abs(t) t.

    3) If f(x) = x^2 6x + 5, sketch a graph of the region defined by the intersection of the inequalities f(x) + f(y) ≤ 0 and f(x) f(y) ≥ 0.

    Could you please help me out?

    Thanx
    The first two are very straightforward. The third - I'm still thinking about!

    (1) Between -1 and 1, the graph is the line y = x; i.e. at 45 degrees to the axes between (-1,-1) and (1, 1), excluding (-1, -1), but including (1, 1). This covers a range of values of x of length 2, the period of the function. So this line segment is repeated between x = 1 and x = 3, and again, and again, ... Also to the left between x = -3 and x = -1, etc... See sketch.

    (2) Assuming t is real, the domain is all the real numbers.

    If t \ge 0, |t| = t \Rightarrow |t| - t = 0

    If t < 0, |t| = -t \Rightarrow |t|-t = -2t which is > 0 for t < 0.

    So the range is all the non-negative reals.

    Grandad
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  3. #3
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    Hmmmm perhaps I have to shade the regions f(x) ≤ - f(y) and f(x) ≥ f(y)</SPAN>

    BUT I have no idea what the final image should look like (like the final areas shaded)

    Could you please show me?
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  4. #4
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    Hello xwrathbringerx
    Quote Originally Posted by xwrathbringerx View Post
    3) If f(x) = x^2 6x + 5, sketch a graph of the region defined by the intersection of the inequalities f(x) + f(y) ≤ 0 and f(x) f(y) ≥ 0.

    Could you please help me out?

    Thanx
    Sorry I've been slow in 'thinking'!

    f(x) + f(y) = x^2-6x+5+y^2-6y+5= (x-3)^2 + (y-3)^2 -8

    So f(x) + f(y) = 0 is a circle, centre (3,3) radius \sqrt8\, (= 2\sqrt2).

    And f(x) + f(y) \le 0 represents the points on and inside this circle.

    f(x) - f(y) = (x^2 -6x)-(y^2-6y) =(x-3)^2-(y-3)^2, which is a hyperbola, centre (3,3) with asymptotes parallel to x = \pm y (Do you know what this looks like?)

    So f(x) - f(y) \ge 0 represents the points on and 'outside' this hyperbola (i.e. points on the far side of the curve from the centre).

    Can you sketch these now?

    Grandad
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