What is the value of the local linearization of the function $f\left(x\right)=x^2$ at $x=2$, evaluated at $x=2.1$?

• A. $4.2$
• B. $4.05$
• C. $4.41$
• D. $4.0$

Answer: (A)

Verified step by step guidance

1. Step 1: Recall the formula for local linearization, which is given by L(x) = f(a) + f'(a)(x - a), where 'a' is the point of tangency, 'f(a)' is the value of the function at 'a', and 'f'(a)' is the derivative of the function evaluated at 'a'.
2. Step 2: Compute f(a) by substituting a = 2 into the function f(x) = x^2. This gives f(2) = 2^2 = 4.
3. Step 3: Find the derivative of the function f(x) = x^2. The derivative is f'(x) = 2x. Evaluate f'(a) at a = 2, which gives f'(2) = 2(2) = 4.
4. Step 4: Substitute the values f(a) = 4, f'(a) = 4, and x = 2.1 into the linearization formula L(x) = f(a) + f'(a)(x - a). This gives L(2.1) = 4 + 4(2.1 - 2).
5. Step 5: Simplify the expression L(2.1) = 4 + 4(0.1) to find the approximate value of the function at x = 2.1 using local linearization.