If \(\rm f(x) = \left \{ \begin{matrix} \rm x^2; & \rm x \leq 0 \\ \rm 2\sin x; & \rm x > 0 \end{matrix}\right.\), then x = 0 is a point of:

Concept:

Continuity of a Function:

• A function f(x) is said to be continuous at a point x = a in its domain, if \(\rm \displaystyle \lim_{x\to a}f(x)\) exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ \(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).

Differentiability of a Function:

• A function f(x) is differentiable at a point x = a in its domain if its derivative is continuous at a.

  This means that f'(a) must exist, or equivalently: \(\rm \displaystyle \lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=\lim_{x\to a}f'(x)=f'(a)\).

Maxima/Minima:

• If f(x) has a local maximum or a local minimum at a point x = a, then it must be either a critical point [f'(a) = 0] or a point of non-differentiability.

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Calculation:

Let us check for the continuity and differentiability (maxima/minima) of the function at x = 0.

Continuity:

\(\rm f(x) = \left \{ \begin{matrix} \rm x^2; & \rm x \leq 0 \\ \rm 2\sin x; & \rm x > 0 \end{matrix}\right.\)

\(\rm \displaystyle \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}x^2\) = 0² = 0.

\(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}2\sin x\) = 2 sin 0 = 0.

f(0) = 0² = 0.

∵ \(\rm \displaystyle \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)\) = f(0), the function f(x) is continuous at x = 0.

Differentiability:

\(\rm f'(x) = \left \{ \begin{matrix} \rm 2x; & \rm x \leq 0 \\ \rm 2\cos x; & \rm x > 0 \end{matrix}\right.\)

\(\rm \displaystyle \lim_{x\to 0^-}f'(x)=\lim_{x\to 0^-}2x\) = 2×0 = 0.

\(\rm \displaystyle \lim_{x\to 0^+}f'(x)=\lim_{x\to 0^-}2\cos x\) = 2 cos 0 = 2.

∵ \(\rm \displaystyle \lim_{x\to 0^+}f'(x)\neq\lim_{x\to 0^-}f'(x)\), the function is not differentiable at x = 0.

Since, the function is not differentiable at x = 0, let us examine the possibility of maximum/minimum at the point.

The function f(x) = x² is strictly decreasing in (-∞, 0] and its minimum is 0² = 0 at x = 0.

The function f(x) = 2sin x is strictly increasing in \(\rm \left(0,\dfrac{\pi}{2}\right]\) and its minimum is 2 sin 0 = 0 at x = 0.

∴ The function f(x) has a local minimum at x = 0.