Find the local linear approximation of the function

Question

Find the local linear approximation of the function

f(x) = sqrt(7+x)

at x₀ = 2

and use it to approximate: sqrt(8.9) and sqrt(8.1)

a)f(x) = sqrt(7+x) = ___

b) sqrt(8.9) = ___

c) sqrt(9.1) = ___

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.

Answer 1

Hi Kate,

Linear approximation is a method which uses the tangent line to a curve at a known point to approximate the values of unknown points nearby.

Your function is

f(x) = sqrt(x + 7)

The point around which you'll be making your approximations is at x = 2

f(2) = sqrt(2 + 7) = sqrt(9) = 3

So you'll be using the known point (2,3) for your approximations.

You need to make a tangent line to the curve sqrt(x + 7) at the point (2,3) in order to do your approximation.

Start by taking the derivative:

f'(x) = 1/2*sqrt(x + 7)^-1/2

Plug in x = 2

f'(2) = 1/2*sqrt(2 + 7)^-1/2

f'(2) = 1/6

This will be the slope of your tangent line. Since you also know the point, you can use point-slope form to draw your line.

y - y1 = m(x - x1)

y - 3 = 1/6*(x - 2)

This is your answer for a.

b) To get sqrt(8.9) from the function sqrt(x + 7), you must have plugged in x = 1.9. Plug this into your linear approximation.

y - 3 = 1/6*(1.9 - 2)

y - 3 = -0.0167

y = 2.983333

c) To get sqrt(9.1) from the function sqrt(x + 7), you must have plugged in x = 2.1. Plug this into your linear approximation.

y - 3 = 1/6*(2.1 - 2)

y - 3 = 0.0167

y = 3.016667

The actual answers for sqrt(8.9) and sqrt(9.1) are 2.983287 and 3.016621. We slightly over approximated our answers because the tangent line sits above the curve in this situation.

Hope this helps.