Find the local linear approximation of the function f(x) = √(16 + x) at x0 = 0, and use it to approximate √15.9 and √16.1.

Question

Find the local linear approximation of the function f(x) = √(16 + x) at x₀ = 0, and use it to approximate √15.9 and √16.1.

(a) f(x) = √(16 + x) ≈ 4 + (1/8)x

(b) √15.9 ≈

(c) √16.1 ≈

For parts (b) and (c), you should enter your answer as a fraction. If you enter a decimal, make sure that it is correct to at least six decimal places.

Let y = 2x^2.

Find the change in y, Δy when x = 4 and Δx = 0.4

Answer 1

If f(x) = √(16 + x) or f(x) = (16 + x)^1/2 then f '(x) = 1/2(16 + x)^-1/2 = 1/(2√(16 + x))

The point-slope form of a line is y - y₁ = m(x - x₁) and we are are going to solve for y₁ (aka, a new "y"):

y₁ = y - m(x - x₁) but we are going to use the derivative as "m":

y₁ = y - f '(x)(x - x₁)

y₁ is our "new y" or the estimation of the function value when we plug in x₁ into our original function.

In our problem, if we assume x₀ = 0 that is what we will plug into f(x) and f '(x)

To find √15.9, we would need to have x₁ = -0.1 since the function is √(16 + x) to get √15.9 requires x = -0.1.

f(0) = √(16 + 0) = 4 so we let y = 4

f '(0) = 1/(2√(16 + 0)) = 1/8 = 0.125

So, plugging those in we get:

y₁ = 4 - 0.125(0 - (-0.1)) = 4 - 0.125(0.1) = 3.9875 = 319/80

So, our approximation of √15.9 is 3.9875 or 319/80

To find √16.1, we would need to have x₁ = 0.1 since the function is √(16 + x) to get √16.1 requires x = 0.1.

y₁ = 4 - 0.125(0 - (0.1)) = 4 - 0.125(-0.1) = 4.0125 = 321/80

So, our approximation of √16.1 is 4.0125 or 321/80

For y = 2x², y' = 4x meaning dy/dx = 4x

Since dy/dx = 4x then, by multiplying "dx" on both sides we get

dy = 4x(dx) or (using Δ's):

Δy = 4x(Δx)

So when x = 4 and Δx = 0.4, Δy (or the change in y) = 4(4)(0.4) = 6.4

Notice that this is just an approximation of the change in y. The ACTUAL change in y is y(4.4) - y(4) = 2(4.4)² - 2(4)² = 6.72