Question

The equation of the tangent line to f(x) = \sqrt{x} at x = 49 is y = .
Using this, we find our approximation for \sqrt {49.3} is

Answer 1

2 separate problems

Answer 2

Well first you need to find the tangent line of f(x)=sqrt(x) at x=49

Answer 3

how do you do that

Answer 4

\[y-f(49)=f'(49)(x-49) \\ y=f'(49)(x-49)+f(49)\]

Answer 5

that is point slope form above

Answer 6

we just need to know the derivative of f
and then plug in 49 into it
we also need to know f(49) too

Answer 7

I don't understand

Answer 8

which part?

Answer 9

finding f'?

Answer 10

ok i guess you got it since you closed it