Equation of a line tangent to...

[ladyfire]

find an equation of the line tangent to the curve y= x+ (1/x) at the point (5, 26/5)


~okay so I use the equation f(x+h) -f(x) / h
~I then plug in my x, giving me 5 +h / h
~then I plug it into my original equation y=x+(1/x), giving me y= (5+h) + (1/ (5+h))
~Since I am adding, I need common denomenators. so I multiply the (5+h)/1 by (5+h) to get [(5+h)(5+h) / (5+h)] + 1/(5 +h)

~I decide to leave my equation as thus:

(5+h)(5+h)+1
_____________
(5+h)


should I cancel out one (5+h) or should I factor out the h to get h(5+1)(5+1)+1?

do you end up with y= 6 or 1?

then what do i do?

 

[skeeter]

to find the value of f'(5) for the function f(x) = x + 1/x using one of the definitions of a derivative, then proceed as follows:

f'(5) = lim{x->5} [f(x) - f(5)]/(x - 5)

f'(5) = lim{x->5} [x + 1/x - 26/5]/(x - 5)

f'(5) = lim{x->5} (5x² - 26x + 5)/[5x(x - 5)]

f'(5) = lim{x->5} (5x - 1)(x - 5)/[5x(x - 5)]

f'(5) = lim{x->5} (5x - 1)/(5x) = 24/25

now you have the slope and point ... write the tangent line equation.