Understanding the Derivative of a Piecewise Continuous Function at a Non-Differentiable Point
When dealing with functions that are continuous but not differentiable at certain points, understanding their behavior can be quite intricate. In this article, we explore the derivative of a product function at a specific point, specifically where the function is continuous but not differentiable. We'll look at the case of finding the derivative of f(x) xg(x) at x0, where g(0) 8, while g(x) is continuous but not differentiable at x0.
Concept and Definitions
To solve for f(0), we can utilize the product rule for differentiation. The product rule states:
if u(x) x and v(x) g(x), then:
f(x) u(x)v(x) u'(x)v(x)
Step-by-Step Derivation
Let's break down the steps to evaluate f(0).
Step 1: Define u(x) and v(x)
In our case:
u(x) x v(x) g(x)Step 2: Evaluate u'(x) and u(x) at x0
We know:
u(x) x implies u'(x) 1 u(0) 0Step 3: Evaluate v(x) at x0
We are given that:
v(0) g(0) 8
Step 4: Evaluate u'(0)
We already found:
u'(0) 1
Step 5: Use the Product Rule to Find f(0)
Using the product rule:
f(0) u(0)v(0) u'(0)v(0)
Substituting the values:
f(0) (0)(8) (1)(8) 0 8 8
Note on v(0)
Even though g(x) is not differentiable at x0, we can still evaluate v(0) g(0) 8. The critical factor here is that u(0) 0, which makes the second term of the product rule zero, hence the only contributing term is the first term.
Conclusion
The evaluation demonstrates that:
f'(0) 8
Therefore, the final answer is:
boxed{8}
Additional Insights
In this scenario, the continuity of g(x) at x0 allows us to evaluate the limit of f(x) as x approaches 0:
lim_{x to 0} frac{f(x) - f(0)}{x - 0} lim_{x to 0} frac{x g(x) - 0}{x} lim_{x to 0} g(x) 8
This confirms our solution and underscores the importance of continuity in certain contexts even when differentiability is lacking.