1. range of values

I can't remember how to find the range of values of a function to save my life!

e.g.
Given that 2x+3y-5=0, find the range of values of x if y<6.

2. Originally Posted by colloquial
I can't remember how to find the range of values of a function to save my life!

e.g.
Given that 2x+3y-5=0, find the range of values of x if y<6.

One way:
$2x + 3y - 5 = 0$

So
$y = -\frac{2}{3}x + \frac{5}{3}$

So you want to solve
$-\frac{2}{3}x + \frac{5}{3} < 6$
for x.

-Dan

3. also the question:
Find the range of values of p if 3p^2 + 2p >= 5
so I did:
3p^2 + 2p - 5 >=0
(3p+5) (p-1)>=0
is that even correct?
:S

then I would have to draw the number line right?

thanks again!

4. Originally Posted by colloquial
also the question:
Find the range of values of p if 3p^2 + 2p >= 5
so I did:
3p^2 + 2p - 5 >=0
(3p+5) (p-1)>=0
is that even correct?
:S

then I would have to draw the number line right?

thanks again!
Here you need to be a bit more careful.

$3p^2 + 2p \geq 5$

$3p^2 + 2p - 5 \geq 0$

$(3p + 5)(p - 1) \geq 0$

Now split the number line into 3 pieces: $\left ( -\infty, -\frac{5}{3} \right ), \left ( -\frac{5}{3}, 1 \right ), (1, \infty)$
(based on where the LHS is 0) and test your inequality on each of these intervals. The intervals that work give you your solution set.

In this case:
$\left ( -\infty, -\frac{5}{3} \right )$: $(3p + 5)(p - 1) \geq 0$

$\left ( -\frac{5}{3}, 1 \right )$: $(3p + 5)(p - 1) \leq 0$

$(1, \infty)$: $(3p + 5)(p - 1) \geq 0$

So your solution set will be
$\left ( -\infty, -\frac{5}{3} \right ] \cup [1, \infty)$.
(I have included the end points because of the $\geq$ relation. If this had been a > you would not include these points.)

-Dan