Linear Approximation?

Question

Find the local linear approximation of the function
y= sqr(1+x) at x_0=0, and use it to approximate sqr0.9 and sqr1.1.

I got 1+1/2x for the linear function, but i didn't get the approximate number.

Answer 1

y = (1 + x)^½
≈ 1 + ½x
So 1.1^½ = (1 + 0.1)½
≈ 1 + ½*0.1
= 1.05 (Note 1.05*1.05 = 1.1025 .. not bad!!)

0.9^½ = (1 - 0.1)*½
= (1 + (-0.1))^½
≈ 1 + ½*-0.1
= 0.95 (Note 0.95*0.95 = 0.9025 .. also not bad!!)

Answer 2

I answered this earlier. The linear approximation is

L(x) = f(x0) + f'(x0)(x-x0),

where f(x) =sqrt(1+x), x0=0, and x=0.1 or -0.1.

You get L(x) = 1 + (1/2)x, so plug in x=0.1 to get 1.05, and x=-0.1 to get 0.95. The exact square roots are 1.0488 and 0.9487, so these are decent approximations.

Answer 3

y = (1+x)^(1/2)
dy/dx = (1/2)(1+x)^(-1/2)
@ x = 0,y = 1, y' = 0.5
0.5 = (y-1)/(x-0)
y - 1 = 0.5x
y = 0.5x + 1

0.9 = 1 + (-0.1)
y(-0.1) = -0.05 + 1 = 0.95, (0.95)^2 = 0.9025
1.1 = 1 + 0.1
y(0.1) = 0.05 + 1 = 1.05, (1.05)^2 = 1.1025