## anonymous one year ago If g(x) = sqrt(x-3) and k(x) = x^2 + 5 then determine (kg)(x).

1. rhr12

(k-g)(5) means plug 5 into both equations, then divide the answer of k(x) by g(x) (g o k)(7) means plug 7 the equation k(x) and then plug the answer you get from k(x) into g(x)

2. rhr12

(k - g)(5) =(x²+5)–x–3 = x²-x+2 = 5²-5+2 = 22

3. anonymous

$(kg)(x)=(x^2+5)(\sqrt{x ^{-3})}$

4. rhr12

hope it helps

5. anonymous

Where did you get (k-g) @rhr12? It says to multiply k and g. Were you just giving an example?

6. anonymous

I agree with saseal that $(kg)(x) = k(x)g(x) = (x^2+5)\sqrt{x-3}$

7. anonymous

So what would you plug in for x?

8. anonymous

Or would you have to multiply that first?

9. anonymous

$\sqrt{x-3}=(x-3)^{\frac{ 1 }{ 2 }}$

10. anonymous

theres nothing to plug in, theres no value given in the question

11. anonymous

So I have to multiply it....Couldn't I just square both equations to get rid of the square root?

12. anonymous

i dont think you can multiply, just leave it that way

13. anonymous

I don't know think I'd bother multiplying it out, if its actually possible. If I wanted to evaluate (kg) at say x=10, I'd just use it as a normal function and substitute x = 10 everywhere. i.e.$(kg)(10) = (10^2+5)\sqrt{10-7} = 105\sqrt{7}$

14. anonymous

So just write (x^2 + 5)(sqrt(x - 3))?

15. anonymous

yea

16. anonymous

It should be $\sqrt{10-3} \text{ instead of } \sqrt{10-7}$ and yep that is what I'd write

17. anonymous

Alright thanks!