Sketching graphs

• Aug 20th 2009, 12:07 AM
xwrathbringerx
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.

Thanx
• Aug 20th 2009, 05:11 AM
Hello xwrathbringerx
Quote:

Originally Posted by xwrathbringerx
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.

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.

• Aug 25th 2009, 02:12 AM
xwrathbringerx
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)

• Aug 25th 2009, 04:19 AM
Hello xwrathbringerx
Quote:

Originally Posted by xwrathbringerx
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.

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?