SOLUTION: In consider the function 1/x
Find the slope of the line L tangent to the graph of (3,f(3))
I know to find the slope of the tangent I use: lim h->0 f(x+h)-f(x)/h
why d
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-> SOLUTION: In consider the function 1/x
Find the slope of the line L tangent to the graph of (3,f(3))
I know to find the slope of the tangent I use: lim h->0 f(x+h)-f(x)/h
why d
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Find the slope of the line L tangent to the graph of (3,f(3))
I know to find the slope of the tangent I use: lim h->0 f(x+h)-f(x)/h
why does this switch from that to (1/(x+h))-(1/x)/(h)?
I thought since we have x and y values (3,f(3)) I can just somehow plug those values in. Why does x+h become 1/(x+h)
Do we first have to alter the limit definition of a derivative (tangent slope) first given the graph shape? then do some rearranging and factoring? Because I know it gets to be lim h->0 1/x^2 which makes the slope m = -1/9
what I also don't get is why do y become y=1/3 when the original points were (3,f(3))
Confused so much about all this.
Please help
Thank you Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! We are working with the limit definition of the first derivative, that is
Limit as h --》0 of ( f(x+h) - f(x) ) / h
We are given f(x) = 1 / x
We apply the given function to the limit definition
Limit as h --》0 of ( 1/(x+h) - 1/x ) / h =
( x - x - h ) / (x(x+h)h ) =
-h / (x^2+xh)h =
-1 / x^2
Note that the xh term becomes 0 as h --》0
f(3) = 1 / 3
m = f'(3) = -1 / 3^2 = -1 / 9
