Question

1.Let g(x) =root(x^2 + 16 )
a.Find the linearization of g(x) at x = 3.
b.Use your linearization to approximate g(2.9).

Answer 1

Hay are you still there???

Answer 2

yes

Answer 3

ok I do not know what to do sorry lol but if you want you could give me a medal for trying lol PLz ^.^ Thanks

Answer 4

Can you please @butterflyprincess ????

Answer 5

use the formula L(x) = g(a) + g'(a)(x-a)

part a: when they say at x=3 they want the tangent line at the point (3, 5)
so its a = 3 when finding the linearization
\[g(3) = \sqrt{3^2 + 16} = 5\]
g'(x) = \[\frac{ 1 }{ 2 } (x^2 + 16)^\frac{ -1 }{ 2 } (2x) = \frac{ x }{ \sqrt{x^2 + 16} }\]
so g'(3) = 3/5

so now you have L(x) = 5 + 3/5(x-3)
L(x) = 5 + 3/5x - 9/5
\[L(x) = \frac{ 16 }{ 5 } + \frac{ 3 }{ 5 }x \]

now for part b:
to estimate g(2.9) so substitute it in the linearization equation

L(2.9) = 16/5 + 3/5(2.9) = 4.94

you can check this by doing g(2.9) = 4.94064773