To find the linearization, L(x), of the function f(x) at a point a, we follow the formula:
L(x) = f(a) + f'(a)(x – a)
In this case, we have the function f(x) = x^4 + 5 and we want to find the linearization at the point a = 2.
First, we need to evaluate f(a):
f(2) = 2^4 + 5 = 16 + 5 = 21.
Next, we need to find the derivative of f(x):
f'(x) = 4x^3.
Now we evaluate f'(a):
f'(2) = 4(2^3) = 4(8) = 32.
Now, substituting these values back into the linearization formula:
L(x) = f(2) + f'(2)(x – 2)
L(x) = 21 + 32(x – 2).
Therefore, we have:
L(x) = 21 + 32x – 64
L(x) = 32x – 43.
So, the linearization L(x) of the function f(x) = x^4 + 5 at the point x = 2 is:
L(x) = 32x – 43.