## anonymous one year ago Help with difference quotient..

1. anonymous

2. anonymous

@Nnesha

3. Nnesha

substitute x for x+h

4. anonymous

So... f(x)-f(x) / h

5. Nnesha

function is f(x) = sqrt{14x} <--replace x with x+h to find f(x+h)

6. Nnesha

formula is $\large\rm \frac{ f(\color{ReD}{x+h}) -f(x) }{ h }$

7. anonymous

So square root 14(x+h)

8. anonymous

Im at.. square root 14 (x+h-x) over h(square root x+h + square root x)

9. Nnesha

yes right so $\huge\rm \frac{ \sqrt{14(x+h)} - \sqrt{14x} }{ h }$

10. Nnesha

eh eh what nope you can't combine both sqrts

11. anonymous

?

12. Nnesha

hmm how did you get square root x+h + square root x) at the denominator ?

13. anonymous

Final answer--> square root 14 over square root x+h + square root x

14. Nnesha

sqrt{14(x+h) } can be written as sqrt{14} sqrt{x+h) so $\frac{ \sqrt{14}\sqrt{x+h} -\sqrt{14}\sqrt{x}}{ h }$ sqrt 14 is common so you can take it out

15. anonymous

Um not sure where your at

16. Nnesha

$\huge\rm \frac{ \sqrt{14} }{ \sqrt{x+h}+\sqrt{x}}$ is this ur final answer ?

17. anonymous

Yes and its correct.

18. Nnesha

ahh i see so you have to multiply top and bottom with the conjugate of the numerator

19. anonymous

Yup

20. Nnesha

sqrt{14(x+h) } can be written as sqrt{14} sqrt{x+h) so $\huge\rm \frac{ \sqrt{14}\sqrt{x+h} -\sqrt{14}\sqrt{x}}{ h }$ sqrt 14 is common so you can take it out did you understand that step ?

21. anonymous

Yeah it gets moved to the front --- square root 14 ( square root x+h) - square root x over h ... That was the second step

22. anonymous

Thanks for helping.

23. Nnesha

yw.