anonymous
  • anonymous
Determine an equation of the line that is tangent to the graph of f(x) = sqrt (x+1) and parallel to x-6y + 4= 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I found the derivative , which is the slope is 1/ 2 sqrt (x+1) what do i do next?
TuringTest
  • TuringTest
you need to find when that the derivative is the same as the slope of the given line what is the slope of the given line?
lgbasallote
  • lgbasallote
get the derivative of sqrt (x-1)...that's your equation of the tangent

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TuringTest
  • TuringTest
what is the slope of x-6y + 4= 0 ?
TuringTest
  • TuringTest
your derivative is wrong, by the way...
anonymous
  • anonymous
is it.... cuz if it is parallel, i can just plug in that slope to the equation..
anonymous
  • anonymous
how is it wrong...
TuringTest
  • TuringTest
maybe it's just your notation\[\frac d{dx}\sqrt{x+1}=\frac d{dx}(x+1)^{1/2}=\frac12(x+1)^{-1/2}\]
anonymous
  • anonymous
yea..it looks the same i have to solve it by limits
TuringTest
  • TuringTest
we need when that is the same as the slope of the given line. m:\[\frac12(x+1)^{-1/2}=m\]oh you need the definition, eh? ok...
TuringTest
  • TuringTest
\[\frac{\sqrt{x+1+h}-\sqrt{x+1}}h\]\[=\frac{\sqrt{x+1+h}-\sqrt{x+1}}h\cdot\frac{\sqrt{x+1+h}+\sqrt{x+1}}{\sqrt{x+1+h}+\sqrt{x+1}}\]\[=\frac{x+1+h-x-1}{h\sqrt{x+1+h}+\sqrt{x+1}}=\frac{h}{h\sqrt{x+1+h}+\sqrt{x+1}}\]\[=\frac{1}{\sqrt{x+1+h}+\sqrt{x+1}}\]which in the limit is\[\frac1{2\sqrt{1+x}}\]so what have you got for m ?
anonymous
  • anonymous
umm..isnt the answer of the derivative = the slope?
anonymous
  • anonymous
and for the line, i got 1/6 as the slope..it looks wrong...i dont know
TuringTest
  • TuringTest
yes, and we want to know when that slope is the same as the slope of the given line\[\frac1{\sqrt{x+1}}=\frac16\]yes, 1/6 is what I got
anonymous
  • anonymous
so do i find x using 1/2 sqrt (1+x) = 1/6 ?
TuringTest
  • TuringTest
yes, you need to know the point x for the point-slope form when we make our line
anonymous
  • anonymous
oh ok thanks!! i got x= 8 :)
TuringTest
  • TuringTest
so now you use the calculus version of point-slope form:\[y-y_1=f'(x_1)(x-x_1)\]
anonymous
  • anonymous
don't really get it ..
anonymous
  • anonymous
do u find a point and then plug in to y2- y1 /x 2- x1 = m?? i got 6y - x + 3 = 0 which is different from the textbook ans

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