Thread: Algebra Of Function Steps

Hello,
I have a couple of function problems that I have the answers to, but the steps shown in the book are confusing the heck out of me. Perhaps someone may be able to explain it a little more clearer.

For the first one, it is asking me to find: (f+g)(x) and (f+g)(5) and apply them to:

f(x)=3x+1, g(x)=2x-6


Also, to these:

f(x)=x-5, g(x) =3x^2

Thanks!

Originally Posted by annon25

Hello,
I have a couple of function problems that I have the answers to, but the steps shown in the book are confusing the heck out of me. Perhaps someone may be able to explain it a little more clearer.

For the first one, it is asking me to find: (f+g)(x) and (f+g)(5) and apply them to:

f(x)=3x+1, g(x)=2x-6

(f+g)(x) = f(x) + g(x) = (3x+1) + (2x-6) = 5x-5

(f+g)(5) = 5(5)-5 = 20

Also, to these:

f(x)=x-5, g(x) =3x^2

you try this one ...

...

Given two functions , we define a new function denoted by in the following way:



If you understand this definition you'll solve easily the problem.

Edited: Sorry, I didn't see skeeter's post.

I'm given the answer 75, but am coming up with a completely different one. Here are my steps

(x-5)+(3x^2)=3x^2+x-5

(5)(3^2)(5) = 225

(5-5) + ( 3 (5)^2) not (5)(3^2)(5)

Originally Posted by annon25

Hello,
I have a couple of function problems that I have the answers to, but the steps shown in the book are confusing the heck out of me. Perhaps someone may be able to explain it a little more clearer.

For the first one, it is asking me to find: (f+g)(x) and (f+g)(5) and apply them to:

f(x)=3x+1, g(x)=2x-6


Also, to these:

f(x)=x-5, g(x) =3x^2

Thanks!

Think of the distributive rule in terms of (f + g)(x).


(f + g)(x) = f(x) + g(x).


For the second question, we know that f(x) = x - 5 and g(x) = 3x^2.

So, (f + g)(x) = x - 5 + 3x^2.


Now we are told to replace x with 5 as indicated by (f + g)(5).

Let x = 5 and plug into x - 5 + 3x^2.

x - 5 + 3x^2

5 - 5 + 3(5)^2

0 + 3(25)

0 + 75

75 is the second answer.