Continuity check at x = 2

Since $2 \le 2$, we use the first piece: $f(2) = 2^2 + 1 = 5$. For the left-hand limit, as $x \to 2^-$ we also use $x^2+1$, giving $\lim_{x\to 2^-} f(x) = 5$. For the right-hand limit, as $x \to 2^+$ we use $5x-5$, giving $\lim_{x\to 2^+} f(x) = 5(2)-5 = 5$.

Since both one-sided limits equal 5, $\lim_{x\to 2} f(x) = 5$ exists. This matches $f(2) = 5$, so all three conditions of continuity are satisfied at $x=2$. Therefore $f$ is continuous at $x = 2$, and in fact continuous on all of $[0,3]$.

Each piece of $f$ is a polynomial, hence continuous on its domain, and part (a) showed the pieces agree in value and limit at $x=2$, so $f$ is continuous on the entire closed interval $[0,3]$.

Evaluate at the endpoints: $f(0) = 0^2+1 = 1$ and $f(3) = 5(3)-5 = 10$. Since $f$ is continuous on $[0,3]$ and $4$ lies strictly between $f(0)=1$ and $f(3)=10$, the Intermediate Value Theorem guarantees there exists at least one $c$ in $(0,3)$ with $f(c) = 4$.

The answer is ten. In part (b), the right piece five x minus five was used since three is greater than two, giving five times three minus five, which equals ten. This value served as the upper endpoint for applying the Intermediate Value Theorem.