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
ok so whats the slope there
are u are given a point this line passes through
No, that's all i was given.
yes you are given enough information
you are given a function that tells you the slope at every x value u want
so what is he slope at x=6?
you are also told that f(6)=30 meaning this tangent line passes through the point(6,30)
y-yo=m(x-xo) y-30=f'(6)(x-6)
y=30=4(x-6)
y-30=4(x-6)***
y-30=4x-24
y=4x+6
slope 4
now approximate the value at 6.02
this line gives you a linear approximation
:/ how do i do that. idk the equation for f(x)
its good for values very close to 6
y-30=4(6.02-6) y=4*0.02 + 30 = 30.08
oh, thanks!
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