Consider the function f(x) = |x|3, where x is real. Then the function f(x) at x = 0 is A. continuous but not differentiable B. once differentiable but not twice C. twice differentiable but not thrice D. thrice differentiable

continuous but not differentiable
once differentiable but not twice
twice differentiable but not thrice
thrice differentiable

The correct answer is: A. continuous but not differentiable.

The function $f(x) = |x|^3$ is continuous at $x=0$ because the left and right hand limits are equal to the function value at $x=0$. However, the function is not differentiable at $x=0$ because the derivative does not exist at $x=0$.

The derivative of a function at a point is the limit of the difference quotient as the difference in the $x$-values approaches zero. In other words, the derivative of a function at a point is the slope of the line tangent to the function at that point.

The function $f(x) = |x|^3$ is not differentiable at $x=0$ because the line tangent to the function at $x=0$ does not exist. This is because the function has a sharp corner at $x=0$.

A function can be continuous but not differentiable at a point. This happens when the function has a sharp corner at that point.

A function can also be differentiable but not continuous at a point. This happens when the function has a jump discontinuity at that point.

A function can be both continuous and differentiable at a point. This is the most common case.