176k views
5 votes
Determine the values of a and b that would result in the function f(x) being differentiable at x=3 for f(x)=x²+ax+b.

User Bulki
by
7.5k points

1 Answer

4 votes

Final answer:

For the function f(x) = x² + ax + b to be differentiable at x = 3, it must be continuous and its derivative must exist at that point. Since f(x) is a polynomial, it is naturally continuous everywhere, and its derivative, f'(x) = 2x + a, also a polynomial, is continuous everywhere, including at x = 3. Thus, f(x) is differentiable at x = 3 for any real values of a and b.

Step-by-step explanation:

To determine the values of a and b that make the function f(x) = x2 + ax + b differentiable at x = 3, we must ensure that both the function is continuous at x = 3 and its derivative is continuous at x = 3. The function will be continuous if it's defined at that point and the limit as x approaches 3 from both sides is equal to the function's value at x = 3.

First, the function is a polynomial, and polynomials are continuous everywhere, so f will be continuous at x = 3 for any value of a and b.

Secondly, for the derivative to be continuous at x = 3, we need to take the derivative of f(x) and ensure it exists at x = 3. The derivative of f(x) = x2 + ax + b is f'(x) = 2x + a. For f'(x) to exist at x = 3, we don't need to find a specific value because the derivative of a polynomial is also a polynomial, and hence it is continuous everywhere including at x =3.

Therefore, f(x) is differentiable at x = 3 for any real numbers a and b because its derivative exists and it's continuous at that point.

User Ivan Barayev
by
7.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.