Question

Help with difference quotient..

Answer 1

@Nnesha

Answer 2

substitute x for x+h

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

So square root 14(x+h)

Answer 7

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

Answer 8

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

Answer 9

eh eh what nope you can't combine both sqrts

Answer 10

?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Um not sure where your at

Answer 15

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

Answer 16

Yes and its correct.

Answer 17

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

Answer 18

Yup

Answer 19

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 ?

Answer 20

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

Answer 21

Thanks for helping.

Answer 22

yw.