#### Howdoyoumath

##### New member

- Joined
- Jul 5, 2019

- Messages
- 3

(a) (f+g)(3)=

(b) (f−g)(3)=

(c) (fg)(3)=

(d) (f/g)(3)=

Thanks

- Thread starter Howdoyoumath
- Start date

- Joined
- Jul 5, 2019

- Messages
- 3

(a) (f+g)(3)=

(b) (f−g)(3)=

(c) (fg)(3)=

(d) (f/g)(3)=

Thanks

- Joined
- Apr 22, 2015

- Messages
- 1,651

Hello. What kind of help do you need? That is, where did you get stuck?Need help answering!!! …

Are you able to calculate the value of f(3) and the value of g(3) separately? If so, then add those two numbers (for part a). If you don't understand the symbols f(3) and g(3) -- or how to find their values -- please let us know.

(f+g)(3) means f(3) + g(3)

Likewise, (f−g)(3) means f(3) - g(3)

(fg)(3) = f(3) ∙ g(3)

(f/g)(3) = f(3) / g(3)

If you've started, please show the work you've tried so far.

PS: Check out the

- Joined
- Dec 30, 2014

- Messages
- 3,658

f(x) says to subtract 3 from x and then square the result. So f(3) says to subtract 3 from 3 and then square the result.

(a) (f+g)(3)=

(b) (f−g)(3)=

(c) (fg)(3)=

(d) (f/g)(3)=

Thanks

g(x) says to subtract twice x from 7. So g(3) says to subtract twice 3 from 7, that is 7-6.

- Joined
- Jul 5, 2019

- Messages
- 3

I have absolutely no idea about how i would start to answer this question. I'm guessing I need to graph the f(x)=(x−3)^2 and g(x)=7−2x into my calculator to determine the x values for each. For f(x)=(x−3)^2 the x value would be f(3)=(3−3)^2. Which doesn't make any sense because that would make it 0. Now if 0 is the correct answer do I plug that 0 for f into the given solutions likeHello. What kind of help do you need? That is, where did you get stuck?

Are you able to calculate the value of f(3) and the value of g(3) separately? If so, then add those two numbers (for part a). If you don't understand the symbols f(3) and g(3) -- or how to find their values -- please let us know.

(f+g)(3) means f(3) + g(3)

Likewise, (f−g)(3) means f(3) - g(3)

(fg)(3) = f(3) ∙ g(3)

(f/g)(3) = f(3) / g(3)

If you've started, please show the work you've tried so far.

PS: Check out theRead Before Postingnotice, too. Thanks!

(a) (f+g)(3)=

(b) (f−g)(3)=

(c) (fg)(3)=

(d) (f/g)(3)= ???

If so I think i have an idea on how to finish the problem.

- Joined
- Mar 16, 2016

- Messages
- 1,393

From post #4 it is clear that you don't understand function notation.

(a) (f+g)(3)=

(b) (f−g)(3)=

(c) (fg)(3)=

(d) (f/g)(3)=

Thanks

If \(\displaystyle f(x) = (x - 3)^2 \) then

\(\displaystyle f(1) = (1 - 3)^2 = (-2)^2 = 4 \)

\(\displaystyle f(2) = (2 - 3)^2 = (-1)^2 = 1 \)

So.… \(\displaystyle f(3) = ...….\)

In \(\displaystyle f(x) = (x-3)^2\), "f" is the name of the function (subtract 3 and then square) and x is the variable used in the function. x can take any value and f(x) is calculated accordingly.

\(\displaystyle g(x)=7 - 2x\) is a

So \(\displaystyle g(1) = 7 - 2*1 = 5\) etc.

Now what is f(3) and g(3)? (No graphing is needed)

You have been told above that (f+g)(x) = f(x) + g(x), so (f+g)(3) = f(3) + g(3) = ...... + ...... = ..........

Now do a similar thing with the other 3 questions.

- Joined
- Mar 16, 2016

- Messages
- 1,393

See red aboveI have absolutely no idea about how i would start to answer this question. I'm guessing I need to graph the f(x)=(x−3)^2 and g(x)=7−2x into my calculator to determine the x values for each. For f(x)=(x−3)^2 the x value would be f(3)=(3−3)^2. Which doesn't make any sense because that would make it 0. Yes it is 0 ...why doesn't that make any sense? 0 is a number just like any other number

Now if 0 is the correct answer do I plug that 0 for f into the given solutions like You don't plug 0 for f, you plug 0 for f(3) because f(3) =0

(a) (f+g)(3)= f(3) + g(3) = 0 + ……. you finish it now

(b) (f−g)(3)=

(c) (fg)(3)=

(d) (f/g)(3)= ???

If so I think i have an idea on how to finish the problem.

Please post your answers and we will see if you understand.

- Joined
- Jan 27, 2012

- Messages
- 5,046

Yes, f(3)= 0. Why would that not make any sense? Surely you have seen many functions that have a value for some x? Yes, you "plug" that 0 into the given formulas. What did you get for g(3)?For f(x)=(x−3)^2 the x value would be f(3)=(3−3)^2. Which doesn't make any sense because that would make it 0. Now if 0 is the correct answer do I plug that 0 for f into the given solutions

(Your grammar is a little odd. It is not the "x value" that would be 0. The f value is 0 when the x value is 3. And you do not "plug that 0 for f into the given solutions" because there are NO "given solutions! You plug f(3)= 0 into the given formulas in order to get your solutions.)