How do you find the linearization of the function at x = 4, given the function f(x) = x^4 + 5x + 32?

Finding the Linearization of the Function

To find the linearization of a function at a given point, we need to use the formula for the linearization, which is given by:

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

In this case, we will find the linearization L(x) of the function f(x) = x^4 + 5x + 32 at the point a = 4.

First, we need to evaluate the function at x = 4:

  f(4) = 4^4 + 5(4) + 32
       = 256 + 20 + 32
       = 308

So, f(4) = 308.

Next, we compute the first derivative of the function:

  f'(x) = d/dx (x^4 + 5x + 32)
        = 4x^3 + 5

Now, we evaluate the derivative at a = 4:

  f'(4) = 4(4^3) + 5
        = 4(64) + 5
        = 256 + 5
        = 261

So, f'(4) = 261.

Now that we have f(4) and f'(4), we can substitute these values into the linearization formula:

  L(x) = f(4) + f'(4)(x - 4)
       = 308 + 261(x - 4)

Distributing the derivative:

  L(x) = 308 + 261x - 1044
       = 261x - 736

So the linearization of the function f(x) at x = 4 is:

  L(x) = 261x - 736

In conclusion, the linearization of the function f(x) = x^4 + 5x + 32 at the point x = 4 is:

  L(x) = 261x - 736