Functions — Question 2

Question 2 - Solution

Part (a): Continuity at $x = 2$ We must check:

• $\lim_{x \to 2^-} f(x)$

• $\lim_{x \to 2^+} f(x)$

• $f(2)$

For $x < 2$, use $f(x) = x^2 - 4x + 3$: $$\lim_{x \to 2^-} f(x) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1$$

For $x \geq 2$, use $f(x) = \sqrt{x + 2}$: $$\lim_{x \to 2^+} f(x) = \sqrt{2 + 2} = \sqrt{4} = 2$$

Since: $$\lim_{x \to 2^-} f(x) = -1, \quad \lim_{x \to 2^+} f(x) = 2$$ the limit from left and right are not equal.

Conclusion: $f(x)$ is not continuous at $x = 2$

Part (b): Differentiability at $x = 2$

A function must be continuous at a point to be differentiable there. Since $f(x)$ is not continuous at $x = 2$, it cannot be differentiable there.

Conclusion: $f(x)$ is not differentiable at $x = 2$