Destinyyyy
  • Destinyyyy
Help with difference quotient..
Mathematics
schrodinger
  • schrodinger
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Destinyyyy
  • Destinyyyy
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Destinyyyy
  • Destinyyyy
Nnesha
  • Nnesha
substitute x for x+h

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Destinyyyy
  • Destinyyyy
So... f(x)-f(x) / h
Nnesha
  • Nnesha
function is f(x) = sqrt{14x} <--replace x with x+h to find f(x+h)
Nnesha
  • Nnesha
formula is \[\large\rm \frac{ f(\color{ReD}{x+h}) -f(x) }{ h }\]
Destinyyyy
  • Destinyyyy
So square root 14(x+h)
Destinyyyy
  • Destinyyyy
Im at.. square root 14 (x+h-x) over h(square root x+h + square root x)
Nnesha
  • Nnesha
yes right so \[\huge\rm \frac{ \sqrt{14(x+h)} - \sqrt{14x} }{ h }\]
Nnesha
  • Nnesha
eh eh what nope you can't combine both sqrts
Destinyyyy
  • Destinyyyy
?
Nnesha
  • Nnesha
hmm how did you get square root x+h + square root x) at the denominator ?
Destinyyyy
  • Destinyyyy
Final answer--> square root 14 over square root x+h + square root x
Nnesha
  • 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
Destinyyyy
  • Destinyyyy
Um not sure where your at
Nnesha
  • Nnesha
\[\huge\rm \frac{ \sqrt{14} }{ \sqrt{x+h}+\sqrt{x}}\] is this ur final answer ?
Destinyyyy
  • Destinyyyy
Yes and its correct.
Nnesha
  • Nnesha
ahh i see so you have to multiply top and bottom with the conjugate of the numerator
Destinyyyy
  • Destinyyyy
Yup
Nnesha
  • 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 ?
Destinyyyy
  • Destinyyyy
Yeah it gets moved to the front --- square root 14 ( square root x+h) - square root x over h ... That was the second step
Destinyyyy
  • Destinyyyy
Thanks for helping.
Nnesha
  • Nnesha
yw.

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