SOLUTION: Hello can someone please try and solve these problems, i would greatly appericiate it!! J. Rivera. 1) State the domain of the following: a){{{g(x)=sqrt

Algebra ->  Human-and-algebraic-language -> SOLUTION: Hello can someone please try and solve these problems, i would greatly appericiate it!! J. Rivera. 1) State the domain of the following: a){{{g(x)=sqrt       Log On


   

Question 125279: Hello can someone please try and solve these problems, i would greatly appericiate it!!
J. Rivera.
1) State the domain of the following:
a)g%28x%29=sqrt+sign+x%2B4
Answer:
b)g%28x%29=2x%2B1%2Fx-3+
2x+1 OVER x-s
Answer:
c) h(x)=3x^2+5x-3
Answer:
d) l(x)-2x+3
Answer:
e) m(x)=3/x^2-7 m "3 OVER x^2-7"
Answer:
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Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
1) State the domain of the following:
a)g%28x%29=sqrt+%28+x%2B4%29
Answer:
you know that x%2B4 has to be positive
Therefore if x+%3C+-4 the result would be a negative+number inside the radical which is not allowed.
so, the domain would be -4 to %2B+infinity (0 is allowed)
Interval notation [-4, infinity)

b)g%28x%29=2x%2B1%2Fx-3+
Answer:
denominator cannot be equal to 0; so, if x-3=0 we will have to divide by 0 which doesn't make a sense.
if x-3=0 ...then x=3
the domain would be +-+infinity to 3and from 3 to %2B+infinity
Interval notation (+-+infinity, 3] U (+3, infinity]



c) h(x)=3x^2+5x-3
Answer:
first find square roots
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 3%2Ax%5E2%2B5%2Ax-3=0 ( notice a=3, b=5, and c=-3)





x+=+%28-5+%2B-+sqrt%28+%285%29%5E2-4%2A3%2A-3+%29%29%2F%282%2A3%29 Plug in a=3, b=5, and c=-3




x+=+%28-5+%2B-+sqrt%28+25-4%2A3%2A-3+%29%29%2F%282%2A3%29 Square 5 to get 25




x+=+%28-5+%2B-+sqrt%28+25%2B36+%29%29%2F%282%2A3%29 Multiply -4%2A-3%2A3 to get 36




x+=+%28-5+%2B-+sqrt%28+61+%29%29%2F%282%2A3%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-5+%2B-+sqrt%2861%29%29%2F%282%2A3%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-5+%2B-+sqrt%2861%29%29%2F6 Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


x+=+%28-5+%2B+sqrt%2861%29%29%2F6 or x+=+%28-5+-+sqrt%2861%29%29%2F6



Now break up the fraction



x=-5%2F6%2Bsqrt%2861%29%2F6 or x=-5%2F6-sqrt%2861%29%2F6



Simplify



x=-5%2F6%2Bsqrt%2861%29%2F6 or x=-5%2F6-sqrt%2861%29%2F6



So the solutions are:

x=-5%2F6%2Bsqrt%2861%29%2F6 or x=-5%2F6-sqrt%2861%29%2F6




the domain would be +-+infinity to .47and from .47 to %2B+infinity
Interval notation (+-+infinity, .47] U [+.47, infinity)
or
the domain would be +-+infinity to -2.14and from -2.14 to %2B+infinity
Interval notation (+-+infinity, -2.14] U [+-2.14, infinity)

d) l(x)= -2x+3

Answer:
(+-+infinity, +infinity)
e) m(x)=3/x^2-7 m "3 OVER x^2-7"
Answer:
x%5E2-7 cannot be equal to 0; x%5E2-7=0 ...x%5E2=7 ......x=sqrt%287%29+
the domain would be +-+infinity to +sqrt%287%29and from +sqrt%287%29 to %2B+infinity
Interval notation (+-+infinity, +sqrt%287%29] U [+sqrt%287%29, infinity)