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Mathematics 20 Online
OpenStudy (lena772):

f ' (x) for a given function is (x^2)/(x+3) and we are told that f (6) = 30. Write the equation of the tangent line to f(x) at x = 6 and use the tangent line equation to approximate the value of f (6.02). 29.92 30.02 30.08 34.00 none of these

OpenStudy (dan815):

ok so whats the slope there

OpenStudy (dan815):

are u are given a point this line passes through

OpenStudy (lena772):

No, that's all i was given.

OpenStudy (dan815):

yes you are given enough information

OpenStudy (dan815):

you are given a function that tells you the slope at every x value u want

OpenStudy (dan815):

so what is he slope at x=6?

OpenStudy (dan815):

you are also told that f(6)=30 meaning this tangent line passes through the point(6,30)

OpenStudy (dan815):

y-yo=m(x-xo) y-30=f'(6)(x-6)

OpenStudy (lena772):

y=30=4(x-6)

OpenStudy (lena772):

y-30=4(x-6)***

OpenStudy (lena772):

y-30=4x-24

OpenStudy (lena772):

y=4x+6

OpenStudy (lena772):

slope 4

OpenStudy (dan815):

now approximate the value at 6.02

OpenStudy (dan815):

this line gives you a linear approximation

OpenStudy (lena772):

:/ how do i do that. idk the equation for f(x)

OpenStudy (dan815):

its good for values very close to 6

OpenStudy (dan815):

y-30=4(6.02-6) y=4*0.02 + 30 = 30.08

OpenStudy (lena772):

oh, thanks!

OpenStudy (dan815):

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