Question

Find the equation of the tangent line to the graph of the given function at the point with the indicated x-coordinate

f(x) = x + 3/x ; x=4

Answer 1

@zepdrix

Answer 2

find f(4) and f'(4)....
what u got for these values?

Answer 3

f(4) = 19/4? not sure how to get f'(4)

Answer 4

f(4) is correct...
what is your f'(x) = ???

Answer 5

write out the function this way so u can use the power rule: \(\large f(x) =x+3x^{-1} \)

Answer 6

ok so the derivative is 1+(-3x)?

Answer 7

not quite.... you forgot to subtract 1 from the exponent...

Answer 8

-1+(-3x)

Answer 9

i mean -3x^-2

Answer 10

ok... that's better .. so f'(x) = ???

Answer 11

1+(-3x^-2)

Answer 12

yes... \(\large f'(x)=1-3x^{-2}=1-\frac{3}{x^2} \)
so f'(4) = ????

Answer 13

would i plug in 4 into that equation?

Answer 14

yes.... f'(4) will give you the slope of the tangent line at x=4.

Answer 15

1-3/(4)^2 = 13/16

Answer 16

ok... good...
so you have a point: (4, 19/4)
and a slope: m = 13/16

looks like you can write the equation of the line using point-slope form...

Answer 17

y = 13/16x+19/4?

Answer 18

hang on... my answer is actually in point-slope form...

Answer 19

ok..

Answer 20

i got something different.... :(
try again...

Answer 21

y = 13/16x+4?

Answer 22

this is what i got:
point-slope form: \(\large y-\frac{19}{4}=\frac{13}{16}(x-4) \)
slope-intercept form: \(\large y=\frac{13}{16}x+\frac{3}{2} \)

Answer 23

ok thanks. the slope-intercept form was the one i was looking for

Answer 24

thanks a lot for the help

Answer 25

yw...:)