Author Topic: Questions (Read 1564 times) Tweet Share

Odette · « **on:** May 21, 2008, 08:52:18 pm »

Ok well i seem to have forgotten how to solve composite functions; sad i know but anyways let me get straight to the point.
I'm stuck with the following question f(x)= $\sqrt x$ +1 g(x)= sin(x). Find f $\circ$ f(x)
I tried to do it but I'm getting stuck lol. >.< Please someone help lolz

Glockmeister · « **Reply #1 on:** May 21, 2008, 09:00:50 pm »

Odette · « **Reply #2 on:** May 21, 2008, 09:39:47 pm »

So that's it? And I thought there was something else i could do,(i mean i thought i could simplify it)

Odette · « **Reply #3 on:** May 21, 2008, 09:43:43 pm »

Another quick question,
Is x^2 + 6x + y^2 - 4y + 9 = 0 an ellipse? I think it is but i'm not sure, when i simplified it, it was in the form of an ellipse equation. Can anyone confirm?
Thanks

AppleXY · « **Reply #4 on:** May 21, 2008, 09:51:36 pm »

more like a circle. but circle is a subset of ellipses anyway.

$(x^2+6x+9)-9+(y^2-4y+4)-4+9 = 0$

$(x+3)^{2}+(y-2)^{2} = 2^{2}$

Odette · « **Reply #5 on:** May 21, 2008, 10:06:01 pm »

Quote from: AppleXY on May 21, 2008, 09:51:36 pm

more like a circle. but circle is a subset of ellipses anyway.

$(x^2+6x+9)-9+(y^2-4y+4)-4+9 = 0$

$(x+3)^{2}+(y-2)^{2} = 2^{2}$

Ohhh Thanks =) Lolz -feels dumb-

Anyways another question =)
The line with equation y= c - x just touches the parabola with the equation y = $2x^{2} + 3x -5$ . What are the values of c?
Sorry I'm a little rusty >.<
Thanks again!

dcc · « **Reply #6 on:** May 21, 2008, 10:15:41 pm »

For these two graphs to touch, they must intersect at some point, so consider:

$\implies c - x = 2x^2 + 3x - 5$

$\implies 0 = 2x^2 + 4x - (5 + c)$

Now since these two equations "just touch" (i.e. they only touch once), this quadratic will have only one solution!

Remembering the discriminant for a quadratic equation ( $\Delta = b^2 - 4ac\mbox{ for: }ax^2 + bx + c$ ), the discriminant is zero when a quadratic has one solution, so we can set the discriminant to zero and solve for c:

$\Delta = b^2 - 4ac = (4)^2 - 4(2)(-(5 + c)) = 16 + 8(5 + c) = 16 + 40 + 8c = 56 + 8c$

Setting this to zero:

$0 = 56 + 8c \implies \boxed{c = -7}$

Odette · « **Reply #7 on:** May 21, 2008, 10:23:23 pm »

Thanks dcc =)
Not sure if this should be posted here but anyways another question,(those doing methods dont worry about this question lolz)
A graph has a polar equation r= 3+ sin(2 $\theta$ ). Find a cartesian form for the equation.

Oh another one lolz,
z= $\frac {1 + 2i} {-1 + 2i}$ , find |z + 1| and $\overline {z}$

Thanks!!!

Glockmeister · « **Reply #8 on:** May 21, 2008, 11:28:02 pm »

This is more Spech, but anyway

$r= 3+ sin(2\theta)$ , (are you sure it's not cis?)

$z= \frac {1 + 2i} {-1 + 2i}$

$z = \frac {1 + 2i} {-1 + 2i} \times \frac {-1 - 2i} {-1 - 2i}$ (valid because $\frac {-1 - 2i} {-1 - 2i} = 1$

$z = \frac {(1+2i)(-1-2i)} {5}$

$z = \frac {-1-2i-2i+4} {5}$

$z = \frac {3}{5} - \frac {4i} {5}$

From the definition of the conjugate,

$\overline {z} = \frac {3}{5} + \frac {4i} {5}$

For the second question

Considering $z = \frac {3}{5} - \frac {4i} {5}$

Let $f(z) = z$
Therefore $f(z+1) = z+1$

$|z+1| = \sqrt {(\frac{3}{5} + 1)^{2} + (\frac{4}{5})^{2}}$

$= \sqrt {(\frac{64}{25}) + (\frac{16}{25})}$

$= \sqrt {\frac{80}{25}$

$= \sqrt {\frac{16}{5}$

(Not sure if that's correct, could someone else have a look.

Odette · « **Reply #9 on:** May 21, 2008, 11:37:21 pm »

Nope its not cis

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