Need help on a calculus problem please.?

Question

Find an equation of the line that is tangent to the graph of f and parallel to the given line.

f(x)=1/(root x-1)
Given line : x+2y+7=0

Please help!

The 1 is within the square root
So its square root (x-1)

Oh yeah, another problem I'm having with this problem is the steps shown to find the derivative of the function. The square root on the bottom is throwing me off and I'm not sure how you end up with the derivative.

Since our class hasn't learned the power rule yet, I have to write my work out the long way using f(x+deltax)-f(x)/deltax. I just end up getting stuck with the rationalizing and can't get it to cancel out into something simple. If you can somehow show me the steps you took to get the derivative using that long way I'd appreciate it very much.

Answer 1

**Is it ....1 / sqrt ( x - 1) , or 1 / [ sqrt(x) - 1 ] ?
e.g. is the 1 inside with the sqrt ????

I will assume it is inside, and use 1 / [ sqrt ( x - 1 ) ]
... however, it is written as if the 1 is outside the sqrt.....

to be tangent, find f ' ( x) ....
f ' ( x ) = ( -1 / 2 ) [ x - 1 ]^ (- 3 / 2 )

line is y = - x / 2 - 7 /2 , so slope is m = - 1/2

to be parallel, the slopes of the unknown line must = given line...

- 1/2 = ( - 1/2 ) [ x - 1 ]^ (- 3 / 2 )
[ x - 1 ]^ (- 3 / 2 ) = 1 , solving , x = 2

y = 1 / sqrt ( 2 - 1 ) = 1

thus slope = - 1/2 , and passes thru ( 2, 1)

so line is y - 1 = ( -1/2) ( x - 2 ), or x + 2y = 4
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edit.......

the function is 1 / sqrt[ x - 1 ], so write using exponents...

1 / sqrt [ x - 1 ] = [ x - 1 ] ^ ( - 1/2 ).... that is, x - 1 to the - 1/2 power.

now use power rule .....
(d/dx) [ U^n] = n U ^ ( n - 1) * ( dU / dx)

( - 1/2 ) [ [ x - 1 ] ^ ( ( - 1/ 2 ) - 1 ) ] * deriv [ x - 1 ] / dx,

and deriv [ x - 1 ] / dx = 1

so.... f ' ( x ) = ( - 1 / 2 ) [ x - 1 ]^ ( - 3 / 2 ) ...., e.g., ...

- 1/2 times the - 3/2 power of [ x - 1 ]

Answer 2

First find the slope of the given line. Should be -1/2. This will be the slope of the parallel tangent line. So the task is now to find where a line of slope -1/2 is tangent to the function and where this point of tangency occurs. So take the derivitive, f'(x)= (-1/2) (root x-1)^(2/3). You then need to find x for which x^(2/3)=2. Once you do that, go back to f(x) and find f(x) for that value of x. Then you have a point and the slope of the tangent line.