1. ## [SOLVED] Functions

Given$\displaystyle f(x)= x^2 -2x +7$ and $\displaystyle g(x)= 2x-3$

find a. $\displaystyle (f+g)(2)$

b. $\displaystyle (f.g)(2)$

c. $\displaystyle (f/g)(2)$

d. $\displaystyle (f.g)(x)$

Thanks in advance. I don't even know where to start

2. $\displaystyle (f+g)(2)=f(2)+g(2)$

$\displaystyle (fg)(2)=f(2)g(2)$

$\displaystyle \left(\frac{f}{g}\right)(2)=\frac{f(2)}{g(2)}$

$\displaystyle (fg)(x)=f(x)g(x)$

3. I got it. I just need to substitute 2

4. Originally Posted by princess_21
I got it. I just need to substitute 2
For the parts where you're evaluating at 2, yes, you can plug "2" in for "x", evaluate, and then do the mathematical operations.

For the part where you're "evaluating" at x, you'll need to work with the functions directly. But the operations on functions work the exact same way: Take the formulas, and do the operations with the polynomials or whatever, instead of which the numbers.

In this case, you'll be multiplying the two polynomials to get the answer.

5. ]am I doing it right??

$\displaystyle f(x)=x^2-2x+7$

$\displaystyle g(x)=2x+3$

a. $\displaystyle (f+g)(2)$

$\displaystyle x^2-2x+7+2x-3$

$\displaystyle x^2+4$

$\displaystyle 2^2 +4$

$\displaystyle 8$

b. $\displaystyle (f.g)(2)$

$\displaystyle 2x^3-3x^2-4x^2+6x+14x-21$

$\displaystyle 2x^3-7x^2+20x-21$

$\displaystyle 2(2)^3+20(2)-21$

$\displaystyle 16-28+40-21$

$\displaystyle 7$

c. $\displaystyle (\frac{f}{g})(2)$

$\displaystyle \frac{x^2-2x+7}{2x-3}$

$\displaystyle \frac{4-4+7}{4-3}$

$\displaystyle 7$

d. $\displaystyle (FoG)$ =>the sign looks like a small "o"

do this also means f(g(x)) if it is then I answered. ( i replaced x in f(x) using g(x) is this right??

$\displaystyle f(x)=x^2-2x+7$

$\displaystyle (2x-3)^(2)-2(2x-3)+7$

$\displaystyle 4x^2-16x+22$

am i in the right track?? thanks for all your help

6. Originally Posted by princess_21
$\displaystyle f(x)=x^2-2x+7$

$\displaystyle g(x)=2x+3$

a. $\displaystyle (f+g)(2)$

$\displaystyle x^2-2x+7+2x-3$
Where did the "-3" come from? From what you did later, I will guess that the original function (posted above) is typoed, and that g(x) should be "2x - 3". If so, then your value here is correct.

Originally Posted by princess_21
b. $\displaystyle (f.g)(2)$

...

$\displaystyle 7$
Since f(2) = 7 and g(2) = 1 (assuming the correction above is correct), then f(2)*g(2) = (7)(1) = 7, so your answer is correct.

Originally Posted by princess_21
c. $\displaystyle (\frac{f}{g})(2)$

$\displaystyle \frac{x^2-2x+7}{2x-3}$

$\displaystyle \frac{4-4+7}{4-3}$

$\displaystyle 7$
Since f(2)/g(2) = 7/1 = 7, then your solution value is correct.

Originally Posted by princess_21
d. $\displaystyle (FoG)$ =>the sign looks like a small "o"

do this also means f(g(x))
Yes; the raised "o" indicates the composition of functions. (I will guess that you mean "F" and "f" to mean the same function, as well as "G" and "g".)

Originally Posted by princess_21
$\displaystyle f(x)=x^2-2x+7$

$\displaystyle (2x-3)^(2)-2(2x-3)+7$

$\displaystyle 4x^2-16x+22$
That's exactly right!

7. whoa!! thanks..