Combinations of algebraic functions

[QuantumCurt]

Homework Statement

f(x)= [itex]\frac{1}{x}[/itex] g(x)=2√x

Find (f+g)(x)

Homework Equations

Sorry, first post in the homework help section. What do I put here?

The Attempt at a Solution

I feel like I'm missing something obvious here.

(f+g)(x)=[itex]\frac{1}{x}[/itex] + 2√x
=x([itex]\frac{1}{x}[/itex] + 2√x)
= 1 + 2x√x
= 1 + 2√x[itex]^{3}[/itex]
= 1 + 2x[itex]^{\frac{3}{2}}[/itex]The solution in the back of the book is showing the answer as:

= [itex]\frac{1 + 2x^{\frac{3}{2}}}{x}[/itex]Where is the x in the denominator coming from? I multiplied the entire equation by x to clear the fraction, so the way I'm seeing it, there shouldn't be a denominator, right? What am I missing?

 

[Office_Shredder]

The very first line is the correct solution (which they then put over a common demonimator). You then multiplied by x for no reason on the second line

 

[Dick]

Homework Statement

f(x)= [itex]\frac{1}{x}[/itex] g(x)=2√x

Find (f+g)(x)

Homework Equations

Sorry, first post in the homework help section. What do I put here?

The Attempt at a Solution

I feel like I'm missing something obvious here.

(f+g)(x)=[itex]\frac{1}{x}[/itex] + 2√x
=x([itex]\frac{1}{x}[/itex] + 2√x)
= 1 + 2x√x
= 1 + 2√x[itex]^{3}[/itex]
= 1 + 2x[itex]^{\frac{3}{2}}[/itex]


The solution in the back of the book is showing the answer as:

= [itex]\frac{1 + 2x^{\frac{3}{2}}}{x}[/itex]


Where is the x in the denominator coming from? I multiplied the entire equation by x to clear the fraction, so the way I'm seeing it, there shouldn't be a denominator, right? What am I missing?

You can't just 'multiply by x'. That changes the function. You CAN multiply by x and then divide by x. That's doesn't change anything.

 

[QuantumCurt]

Where is the x[itex]^{3/2}[/itex] part coming from? Do I just multiply the 2√x by x to get that part? How is the x in the denominator under the 2√x getting there?

 

[QuantumCurt]

You can't just 'multiply by x'. That changes the function. You CAN multiply by x and then divide by x. That's doesn't change anything.

Ok, so I worked it out correctly, but I just need to divide the whole thing by x at the end to restore it to the original value?

Thanks for the help.

 

[QuantumCurt]

I can't just put the 2√x over a denominator of x without changing the 1/x part, can I? That still doesn't get the x^(3/2) part anyway. Like I said, I feel like I'm missing something pretty obvious here.

 

[Ray Vickson]

I can't just put the 2√x over a denominator of x without changing the 1/x part, can I? That still doesn't get the x^(3/2) part anyway. Like I said, I feel like I'm missing something pretty obvious here.

You are missing the fact that
[tex] \frac{a}{b} + c = \frac{a + bc}{b}[/tex] Apply this to ##a = 1, \; b = x, \; c = 2\sqrt{x} = 2 x^{1/2}##.

 

[QuantumCurt]

You are missing the fact that
[tex] \frac{a}{b} + c = \frac{a + bc}{b}[/tex] Apply this to ##a = 1, \; b = x, \; c = 2\sqrt{x} = 2 x^{1/2}##.

Ahhh...Ok, that makes sense now. I don't know why that wasn't occurring to me. I'm in trig right now, and we're working through the algebra review section currently, all of it has been incredibly simple, but for some reason this one just wasn't clicking.

Thanks for the help.

 

[QuantumCurt]

So...

(f+g)(x)= [itex]\frac{1}{x}[/itex] + 2[itex]\sqrt{x}[/itex]
= [itex]\frac{1+(x)(2x^{1/2})}{x}[/itex]
= [itex]\frac{1+2x^{3/2}}{x}[/itex]


Right?


I've got to play around some more with all of the entry methods for math symbols here, this is a pretty cool system. I'm sure I'll be using it some more in the future.

 

[Dick]

So...

(f+g)(x)= [itex]\frac{1}{x}[/itex] + 2[itex]\sqrt{x}[/itex]
= [itex]\frac{1+(x)(2x^{1/2})}{x}[/itex]
= [itex]\frac{1+2x^{3/2}}{x}[/itex]


Right?


I've got to play around some more with all of the entry methods for math symbols here, this is a pretty cool system. I'm sure I'll be using it some more in the future.

Right. And, yes, TeX is very nice. I probably should use it more.

 

[Mentallic]

So...

(f+g)(x)= [itex]\frac{1}{x}[/itex] + 2[itex]\sqrt{x}[/itex]
= [itex]\frac{1+(x)(2x^{1/2})}{x}[/itex]
= [itex]\frac{1+2x^{3/2}}{x}[/itex]Right?I've got to play around some more with all of the entry methods for math symbols here, this is a pretty cool system. I'm sure I'll be using it some more in the future.

And a tip for using latex, don't do things like
[itex ]\frac{1}{x}[/itex ] + 2[itex ]\sqrt{x}[/itex ]

Just wrap the entire expression in tags:
[itex ]\frac{1}{x} + 2\sqrt{x}[/itex ]

And I'd also suggest you use tex tags [tex ] ... [/tex ] when you want to post an expression on its own line, and keep the itex tags for when you want to post the expression on the same line as your text. The reason for this is that itex on separate lines will run into each other as you can see in your post. So for example:

Math can be as easy as [itex ]2+2=4[/itex ] but suddenly it can be as hard as solving simultaneous equations [tex ]x+y=2[/tex ] [tex ]2x+y=3[/tex ]

which becomes

Math can be as easy as [itex]2+2=4[/itex] but suddenly it can be as hard as solving simultaneous equations [tex]x+y=2[/tex] [tex]2x+y=3[/tex]

 

[QuantumCurt]

Thanks for the tips. Hopefully this comes out a lot clearer.

[tex] f(x)=\frac{1}{x} \quad g(x)=2\sqrt{x}[/tex]
[tex] (f+g)(x)=\frac{1}{x} + 2\sqrt{x}[/tex]
[tex] =\frac{1+(x)(2x^{1/2})}{x}[/tex]
[tex] =\frac{1+2x^{3/2}}{x}[/tex]

 

[Mentallic]

Much better