Question

at which x - value is f continuous but not differentiable?

Explanation:

Step1: Recall the conditions for continuity and differentiability

A function \(y = f(x)\) is continuous at \(x = c\) if \(\lim_{x
ightarrow c^{-}}f(x)=\lim_{x
ightarrow c^{+}}f(x)=f(c)\). A function is differentiable at \(x = c\) if \(\lim_{h
ightarrow0}\frac{f(c + h)-f(c)}{h}\) exists. Geometrically, a function is not differentiable at a point where there is a corner (a sharp - point change in the graph), a cusp, a vertical tangent, or a discontinuity.

Step2: Analyze each point

• At \(x=a\): The function has a vertical tangent. The left - hand derivative and the right - hand derivative do not exist in the usual sense (the slope of the tangent line approaches \(\pm\infty\)). Also, we assume the function is continuous (if we consider the general graph behavior, but if there was a break, it would not be continuous. Since the problem is about continuous but non - differentiable, we focus on the non - differentiable aspect).
• At \(x = b\): The graph has a corner. The left - hand derivative \(\lim_{h

ightarrow0^{-}}\frac{f(b + h)-f(b)}{h}\) and the right - hand derivative \(\lim_{h
ightarrow0^{+}}\frac{f(b + h)-f(b)}{h}\) are not equal. But we need to check continuity. For a function with a corner (like \(y=|x|\) at \(x = 0\)), \(\lim_{x
ightarrow b^{-}}f(x)=\lim_{x
ightarrow b^{+}}f(x)=f(b)\) (continuous).

• At \(x=c\): If there is a local minimum (a smooth part of the graph), the function is differentiable (the derivative is zero at the local minimum for a smooth function).
• At \(x = d\): If it is a local maximum (a smooth part of the graph), the function is differentiable (the derivative is zero at the local maximum for a smooth function).

Answer:

\(b\)