# Solving an Equation with g(x) and h(x)

• Jaco Viljoen
In summary: The domain of h is the set of all real numbers, same as the domain of g.The range of h, as I understand it, is from 1 to 1.5 (inclusive).You don't have to have a continuous function to have a domain - for example, the domain of the function ##f(x) = {x \text{ if } 0 < x < 1 \atop 0 \text{ otherwise}} ## is the set (0,1).
Jaco Viljoen

## Homework Statement

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

## Homework Equations

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

## 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:
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} ?##

Jaco Viljoen
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.

Jaco Viljoen
RUber,
I have used the curly braces as requested.

Last edited:
Jaco Viljoen said:

## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

Dg is a line (-∞,∞)
Fh is two lines that come to a point (2,1) but Dh (-∞,∞)?(I am uncertain if I am correct)
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.
Jaco Viljoen said:
(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

Jaco Viljoen said:

## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

Dg is a line (-∞,∞)
Fh 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
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.

Jaco Viljoen
RUber said:
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.

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

RUber said:
√2−x)2+x/2
Jaco Viljoen said:
RUber,
I have used the curly braces as requested.

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.

Jaco Viljoen
SammyS said:
It's easy enough to check you solutions to the equation by plugging them into the functions.

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)

Mark44 said:
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.
Hi Mark,
I just selected and quoted,
Not quite sure what happened.

Jaco Viljoen said:
Dg is a line (-∞,∞)
Fh is two lines that come to a point (2,1) but Dh (-∞,∞)?
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.

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?

RUber said:
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.

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

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.

Jaco Viljoen
Jaco Viljoen said:
Ruber,
so what would the domain of h be?
I am confused
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.

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

Jaco Viljoen
Jaco Viljoen said:
On all of these the Domaine is (-∞,∞)

On a follow up question it asks for the domain of g*l Dg*l
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".

Jaco Viljoen said:
Dg(-∞,∞)*Dl(-∞,∞)
so Dg*l(-∞,∞) right?
Please minimze your use of notation. "Dg(-∞,∞)*Dl(-∞,∞)" is meaningless, as far as I can see.

Mark44 said:
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.

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

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?

HallsofIvy said:
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?

Hi Hallsoflvy,
This is the question from my textbook:

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:
You said before the "product of the domains" so I am not convinced that you are reading this problem correctly! Is f*g the product of the two functions or the composition (typically written "f o g"), f(g(x)). The domain of the product is just the intersection of the two separate domains so yes, if the problem really asks for the domain of their product, it is just the domain of g. The domain of f(g(x)) is that subset of the domain of f such that g(x) is in the domain of f.

Jaco Viljoen
That notation is much more clear. And the solution only seems to be missing one thing.

Jaco Viljoen
RUber said:
That notation is much more clear. And the solution only seems to be missing one thing.

Yes cause we can take the sqrt of a 0,
so:

Dg*l=Dg∩Dl={x∈ℝ:x≤2}

Thank you Ruber

HallsofIvy said:
You said before the "product of the domains" so I am not convinced that you are reading this problem correctly! Is f*g the product of the two functions or the composition (typically written "f o g"), f(g(x)). The domain of the product is just the intersection of the two separate domains so yes, if the problem really asks for the domain of their product, it is just the domain of g. The domain of f(g(x)) is that subset of the domain of f such that g(x) is in the domain of f.

Hallsoflvy,
I think we are on the same page.

Domain of (g*l)
Thank you Hallsoflvy

Jaco Viljoen said:
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)
You have written this in a very confusing way. The functions g and l are multiplied to give the new function g*l, but you don't multiply the domains together. The domain of the product of the two functions is the intersection of the domains of the functions, which is what you wrote in a later post, and quoted below.
Jaco Viljoen said:
g(x)=-(1/2)x-3
l(x)=√(2-x)-3

My question was Domain ofg*l i.e.(-∞,∞)*(-∞,∞) = (-∞,∞)?
No.
The domain of g is (-∞,∞), which you have. The domain of l is not the entire real line, though, due to the square root.

Jaco Viljoen said:
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}
Yes, that looks fine.

Say for example that you have some function f that isn't defined at x=5, then no matter whether the other function g is defined at x=5 or not, when you have f*g, x=5 will still be undefined because it wasn't defined in f. You have to take the intersection of the domains, which is just where both domains are defined at the same time. If the domain is not defined in either f or g (or both), then it's not defined in f*g.

Jaco Viljoen

## What is an equation?

An equation is a mathematical statement that shows the relationship between two or more quantities. It consists of variables, constants, and mathematical operations.

## What is g(x) and h(x)?

g(x) and h(x) are functions, which are mathematical rules that take in input values (x) and produce output values. In the context of solving an equation, g(x) and h(x) represent two different functions that are being combined in an equation.

## How do I solve an equation with g(x) and h(x)?

To solve an equation with g(x) and h(x), you will need to use algebraic methods such as combining like terms, distributing, and isolating the variable. You may also need to use properties of functions, such as the inverse function property, to solve the equation.

## What are some tips for solving equations with g(x) and h(x)?

Here are some tips for solving equations with g(x) and h(x):

• Simplify the equation by combining like terms.
• Use inverse functions to undo operations on the variable.
• Eliminate any fractions by multiplying both sides of the equation by the common denominator.
• Check your solution by plugging it back into the original equation.

## What are some common mistakes when solving equations with g(x) and h(x)?

Some common mistakes when solving equations with g(x) and h(x) are:

• Forgetting to distribute when necessary.
• Making a mistake when combining like terms.
• Not isolating the variable on one side of the equation.
• Not checking the solution by plugging it back into the original equation.

• Precalculus Mathematics Homework Help
Replies
5
Views
523
• Precalculus Mathematics Homework Help
Replies
10
Views
1K
• Precalculus Mathematics Homework Help
Replies
6
Views
718
• Precalculus Mathematics Homework Help
Replies
4
Views
898
• Precalculus Mathematics Homework Help
Replies
21
Views
1K
• Precalculus Mathematics Homework Help
Replies
4
Views
764
• Precalculus Mathematics Homework Help
Replies
5
Views
629
• Precalculus Mathematics Homework Help
Replies
4
Views
2K
• Precalculus Mathematics Homework Help
Replies
7
Views
2K
• Precalculus Mathematics Homework Help
Replies
3
Views
1K