Find the linearization l(x) of the function at a. f(x) = x^4/5, a = 32

Solution:

Given, the function f(x) = x^4/5

We have to find the linearization L(x) of the function at a = 32.

Using the formula,

L(x) = f(a) + f’(a)(x - a)

Now, 

f(x) = x^4/5

f(a) = f(32) = (32)^4/5

f(a) = 16

f’(x) = 4/5 x^{4/5 - 1} = 4/5x^-1/5

f’(a) = f’(32) = 4/5(32)^-1/5

= 0.4

Substituting the values of f(a) and f’(a), the function becomes

L(x) = 16 + 0.4(x - 32)

Therefore, the linearization of f(x) = x^4/5, a = 32 is L(x) = 16 + 0.4(x - 32).

---

Find the linearization l(x) of the function at a. f(x) = x^4/5, a = 32

Summary:

The linearization of the function f(x) = x^4/5, a = 32 is L(x) = 16 + 0.4 (x - 32)