Question

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

Answer 1

I found the derivative , which is the slope
is 1/ 2 sqrt (x+1)

what do i do next?

Answer 2

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?

Answer 3

get the derivative of sqrt (x-1)...that's your equation of the tangent

Answer 4

what is the slope of
x-6y + 4= 0
?

Answer 5

your derivative is wrong, by the way...

Answer 6

is it.... cuz if it is parallel, i can just plug in that slope to the equation..

Answer 7

how is it wrong...

Answer 8

maybe it's just your notation\[\frac d{dx}\sqrt{x+1}=\frac d{dx}(x+1)^{1/2}=\frac12(x+1)^{-1/2}\]

Answer 9

yea..it looks the same
i have to solve it by limits

Answer 10

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...

Answer 11

\[\frac{\sqrt{x+1+h}-\sqrt{x+1}}h\]\[=\frac{\sqrt{x+1+h}-\sqrt{x+1}}h\cdot\frac{\sqrt{x+1+h}+\sqrt{x+1}}{\sqrt{x+1+h}+\sqrt{x+1}}\]\[=\frac{x+1+h-x-1}{h\sqrt{x+1+h}+\sqrt{x+1}}=\frac{h}{h\sqrt{x+1+h}+\sqrt{x+1}}\]\[=\frac{1}{\sqrt{x+1+h}+\sqrt{x+1}}\]which in the limit is\[\frac1{2\sqrt{1+x}}\]so what have you got for m ?

Answer 12

umm..isnt the answer of the derivative = the slope?

Answer 13

and for the line, i got 1/6 as the slope..it looks wrong...i dont know

Answer 14

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

Answer 15

so do i find x using 1/2 sqrt (1+x) = 1/6 ?

Answer 16

yes, you need to know the point x for the point-slope form when we make our line

Answer 17

oh ok thanks!!
i got x= 8 :)

Answer 18

so now you use the calculus version of point-slope form:\[y-y_1=f'(x_1)(x-x_1)\]

Answer 19

don't really get it ..

Answer 20

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