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September 2, 2014

September 2, 2014

Posted by **brian** on Thursday, August 7, 2008 at 12:52pm.

f(x)=(3x+1)/(sqrt(x^2+x-2))

and both the domain and range of

g(x)=(5x-3)/(2x+1)

- Math -
**Damon**, Thursday, August 7, 2008 at 4:09pmWell, the domain would be all x except where the denominator is zero or the sqrt is of a negative number so we need to find where the denominator is zero and where x^2 + x - 2 is negative.

let's look at that denominator function inside the radical sign:

d = x^2+x -2

It is a parabola that opens up

first find the zeros

0 = (x-1)(x+2)

so it is zero at x = -2 and x =+1

those points must not be in the domain but also all the points between them are out because the d function dips below zero between -2 and +1

so our domain is -2 > x > +1

-----------------------------

Now for the second one the denominator is zero for x = -1/2 and all real x except x = -1/2 is the domain

To find the range, sketch the function

for x << 0, g(x)--> +5/2

for x >> 0, g(x)--> +5/2

for x = 0, g(x) = -3

now look at where the numerator is zero

like

x = .6, g = 0

now look at points close to x = -.5 where g(x) gets huge

like

x = -.6, g = 30

x = -.4, g = -25

that tells you what happens each side of the singularity at x = -.5

We see that g(x) goes to +oo as x approaches -.5 from the left and g(x) goes to -oo as x approaches -.5 from the right.

Therefore the range of g is from -oo to +oo

- Math -
**alk**, Friday, August 8, 2008 at 12:07ami understood the whole first question and then i understood the domain of the second question. i did get confused though when it came to finding the range because the teacher gave us the answers w/o explanations and it said that the range was all numbers except for 5/2. i'm not sure how she got that.

thanks for all your help!

- Math -

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