Question

Find the linear approximation of the function

f(x) = sqrt of (1 − x) at a = 0. Use L(x) to approximate the numbers sqrt of 0.9 and sqrt of 0.99.

Answer 1

The linear approximation of the function f(x) = sqrt(1 − x) at a = 0 is 1 - x/2. Using this approximation, sqrt(0.9) is approximately 0.55 and sqrt(0.99) is approximately 0.505.

Step-by-step explanation:

To find the linear approximation of the function f(x) = sqrt(1 − x) at a = 0, we can use the formula for linear approximation, which is L(x) = f(a) + f'(a)(x-a).

The derivative of f(x) = sqrt(1 − x) is f'(x) = -1/(2*sqrt(1 − x)).

Since we are approximating at a = 0, we have L(x) = f(0) + f'(0)(x-0) = 1 + (-1/2)(x) = 1 - x/2.

Using this approximation, we can approximate sqrt(0.9) as L(0.9) = 1 - 0.9/2 = 1 - 0.45 = 0.55 and sqrt(0.99) as L(0.99) = 1 - 0.99/2 = 1 - 0.495 = 0.505.

Answer 2

1−x−−−−√≈−12x+1

0.9−−−√=1−0.1−−−−−−√≈−12(0.1)+1

To simplify:

−1/2⋅1/100+1=−1/2⋅1/100+200/200=200−1/200=199/200 or =.995

So yes for sqrt (.99), that is a good approximation.

To write linear approximations:

Our curve should be y=f(x) and substitute x=a for the numbers m

Our formula for the linear equation is y=mx+b

So subtsitute

y′=f′(x)

y=f′(a)x+b

Then find the y-intercept, which is b.

Given: (a, f(a))= point of the line

Then substitute, that would be

f(a)=f′(a)⋅a+b

f(a)−f′(a)⋅a=b

So the tangent line to y=f(x) at x=a

Or the equation we will use for linear approximations is:

y=f′(a)x+f(a)−f′(a)⋅a

y=f′(a)⋅(x−a)+f(a)