# Homework Help: Suppose function g and h is defined as follows: g(x)=-(1/2)x-3, h(x)=√(2-x)2+(1/2)x

1. Jun 29, 2015

### Jaco Viljoen

1. The problem statement, all variables and given/known data
Suppose the functions g and h are defined as follows:
g(x)=-(1/2)x-3
h(x)=√(2-x)2+(1/2)x

Write down Dh, Dg+h and solve the equation (g+h)(x)=0

2. Relevant equations
g(x)=-(1/2)x-3
h(x)=√(2-x)2+(1/2)x
(g+h)(x)=0

3. The attempt at a solution
Dg is a line (-∞,∞)
h(x) is two lines that come to a point (2,1) but Dh (-∞,∞)???(I am uncertain if I am correct)
(g+h)(x)=0
(-(1/2)x-3)+(√{(2-x)2}+(1/2)x)=0
-3+√{(2-x)(2-x)}=0
-3+√{4-2x-2x+x2}=0
-3+√{x2-4x+4}=0
√{x2-4x+4}=3
x2-4x+4=32
x2-4x+4=9
x2-4x+4-9=0
x2-4x-5=0
(x+1)(x-5)=0
x=-1 or x=5

Last edited: Jun 29, 2015
2. Jun 29, 2015

### RUber

Your notation on the square root is unclear use \sqrt{...} to show what is inside the radical.
is $\sqrt(2-x)^2 = \sqrt{(2-x)^2} \text{ or } (\sqrt{(2-x)})^2$ or is it $\sqrt (2-x)^2+x/2 =\sqrt {(2-x)^2+x/2} ?$

3. Jun 29, 2015

### RUber

Based on your work, I think that $h(x)= \sqrt{ (2-x)^2 } + x/2$, in that case, yes it is defined for any real x.
I see that you squared 3 to make a quadratic...this is a good approach to avoid certain complications with absolute values.
It is worth noting that $\sqrt{ a^2} = |a|$, so you could have gone right to |2-x| = 3, which has the same solutions.

4. Jun 29, 2015

### Jaco Viljoen

RUber,
I have used the curly braces as requested.

Last edited: Jun 29, 2015
5. Jun 29, 2015

### Staff: Mentor

What does Fh mean? I am fairly sure that Dg means the domain of g, but it would be clearer to say "domain of g". I am completely at a loss on what Fh means, though.

Also, as RUber points out, this part of the formula for h(x) is not clear: √(2-x)2. If you take the square root first, and then square the result, what you get is different when you do the same operations in the reverse order.

6. Jun 29, 2015

### SammyS

Staff Emeritus
It's nearly impossible to avoid being ambiguous when using that radical symbol, short of writing √((2-x)2) .

I assume you mean that $\displaystyle\ h(x)=\sqrt{(2-x)^2\,}-(1/2)x \ .$

If that's the case, then your domain is correct.

It's easy enough to check you solutions to the equation by plugging them into the functions.

7. Jun 29, 2015

### Jaco Viljoen

I did attempt to do it that way, but when rechecking with my above route got a different answer and thought I was mistaken, but after confirming my initial thinking I will recheck and see if I didn't make another mistake somewhere.

Thank you RUber

8. Jun 29, 2015

### Staff: Mentor

Your quote from RUber is mangled. Here is what he actually wrote:
$h(x)= \sqrt{ (2-x)^2 } + x/2$

Please don't use the fancy fonts. Your quote from RUber actually obscures things rather than making them clearer.

9. Jun 29, 2015

### Jaco Viljoen

Hi Sammy,
I do check my own solution, I just want to get feedback that I am answering the question correctly.(I do interpret the question correctly)

10. Jun 29, 2015

### Jaco Viljoen

Hi Mark,
I just selected and quoted,
Not quite sure what happened.

11. Jun 29, 2015

### RUber

Be careful with your words here. g(x) is linear, Dg is $(-\infty, \infty)$ or $\mathbb{R}$.
Fh has not been defined. h(x) is piecewise linear, with one linear expression defining h(x) for x < 2 and another for h(x): x≥2. The two lines intersect at the point (2,1), so h is a continuous function of x.

12. Jun 29, 2015

### Jaco Viljoen

On all of these the Domaine is (-∞,∞)

On a follow up question it asks for the domain of g*l Dg*l
Dg(-∞,∞)*Dl(-∞,∞)
so Dg*l(-∞,∞) right?

13. Jun 29, 2015

### Jaco Viljoen

Ruber,
so what would the domain of h be?
I am confused

14. Jun 29, 2015

### RUber

Usually that will be the case. When you combine functions with addition or multiplication, the $D(g*l) = Dg \cap Dl$, implying that anything that is in both domains is in the domain of the combined function.
Be careful when you divide one function by another, since there are more opportunities for values to fall out of the domain in that case.

15. Jun 29, 2015

### RUber

I was correcting your wording, not your logic. The domain of h was right as you originally had it. h is defined for all real x.

16. Jun 29, 2015

### WWGD

I am not sure about the notation, but I do appreciate your wearing suit-and-tie to make your post!

17. Jun 29, 2015

### Staff: Mentor

What is l (lower-case L)? The functions involved here are g and h.

You also haven't told us what Fh means.

Specialized notation is useful as a shorthand, provided that both the writer and reader understand what the notation represents, which isn't the case here. If you write "domain of g + h" I understand that you're asked to find the domain of the sum of these functions. When you write Dg + h, this does not at all mean the "domain of g + h".

Please minimze your use of notation. "Dg(-∞,∞)*Dl(-∞,∞)" is meaningless, as far as I can see.

18. Jul 1, 2015

### Jaco Viljoen

Hi Mark,
Dg+h is how it is written in my textbook, I will refrain from using shorthand, Fh was n typo it should be function h(x).

Domain of l was on another question similar to this one, but they asked:
Write down Dg*l so domain of g which is(-∞,∞)multiplied by domain of l which is(-∞,∞) without first calculating (g*l)(x)
g(x)=-(1/2)x-3
l(x)=√(2-x)-3

My question was Domain ofg*l i.e.(-∞,∞)*(-∞,∞) = (-∞,∞)?
Thank you,
Have a great day.

Jaco

19. Jul 1, 2015

### HallsofIvy

I know how to multiply two functions but I have no idea what "domain of g which is(-∞,∞)multiplied by domain of l which is(-∞,∞)" means. How are you "multiplying" two sets? Do you mean the Cartesian product?

20. Jul 1, 2015

### Jaco Viljoen

Hi Hallsoflvy,
This is the question from my text book:

Write down Dg*l without first calculating (g*l)(x)
g(x)=-(1/2)x-3
l(x)=√(2-x)-3

My understanding is:
g(x)=-(1/2)x-3 so the Dg =ℝ(Do you agree with this?) (-∞,∞)

for l(x)=√(2-x)-3 so the Dl={x∈ℝ: x<2} (Do you agree with this?) (-∞,2)

Definition says:
the product f*g is defined by:
(f*g)(x)=f(x)*g(x)
for all x∈Df∩Dg

So:
Dg*l=Dg∩Dl={x∈ℝ:x≤2}

Last edited: Jul 1, 2015