Find the linear and quadratic approximations to the following function about x = 0 :f (x) = (1 + x)5and so find approximations for 1.055

Sure, let's start with the linear approximation.

The linear approximation of a function f(x) at a point x=a is given by the formula:

L(x) = f(a) + f'(a)(x - a)

Here, we are asked to find the linear approximation about x = 0. So, a = 0.

The function is f(x) = (1 + x)^5.

First, we need to find f(0) and f'(0).

f(0) = (1 + 0)^5 = 1

To find f'(x), we need to take the derivative of f(x).

f'(x) = 5(1 + x)^4

So, f'(0) = 5(1 + 0)^4 = 5

Now, we can plug these values into the formula for L(x):

L(x) = 1 + 5(x - 0) = 1 + 5x

So, the linear approximation of f(x) = (1 + x)^5 about x = 0 is L(x) = 1 + 5x.

Now, let's find the quadratic approximation.

The quadratic approximation of a function f(x) at a point x=a is given by the formula:

Q(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2 / 2

Here, we already know that f(0) = 1 and f'(0) = 5.

To find f''(x), we need to take the second derivative of f(x).

f''(x) = 20(1 + x)^3

So, f''(0) = 20(1 + 0)^3 = 20

Now, we can plug these values into the formula for Q(x):

Q(x) = 1 + 5(x - 0) + 20(x - 0)^2 / 2 = 1 + 5x + 10x^2

So, the quadratic approximation of f(x) = (1 + x)^5 about x = 0 is Q(x) = 1 + 5x + 10x^2.

Finally, to find approximations for 1.055, we can plug x = 0.055 into the linear and quadratic approximations:

L(0.055) = 1 + 5*0.055 = 1.275

Q(0.055) = 1 + 50.055 + 100.055^2 = 1.280375

So, the linear approximation for 1.055 is approximately 1.275 and the quadratic approximation is approximately 1.280375.