For what choices of α, β and γ does the function have a second order derivatives at x = 1?

Steps to Solve

1. Continuity at $x = 1$

To ensure the function is continuous at $x = 1$, the two pieces must be equal at that point: $$ \alpha(1)^2 + \beta(1) + \gamma = 1^3 + 2(1) + 3 $$ This simplifies to: $$ \alpha + \beta + \gamma = 6 $$

2. First Derivative Continuity at $x = 1$

Next, we need the first derivatives of both pieces to be equal at $x = 1$. The derivative of the first piece is: $$ f'(x) = 3x^2 + 2 $$ Evaluating at $x = 1$ gives: $$ f'(1) = 3(1)^2 + 2 = 5 $$

For the second piece, the derivative is: $$ f'(x) = 2\alpha x + \beta $$ Evaluating at $x = 1$ gives: $$ 2\alpha + \beta = 5 $$

3. Second Derivative Continuity at $x = 1$

Finally, the second derivatives must be equal at $x = 1$. The second derivative of the first piece is: $$ f''(x) = 6x $$ Evaluating at $x = 1$ gives: $$ f''(1) = 6(1) = 6 $$

For the second piece, the second derivative is: $$ f''(x) = 2\alpha $$ Setting them equal gives: $$ 2\alpha = 6 \implies \alpha = 3 $$

4. Substituting for $\alpha$

Substituting $\alpha = 3$ into the equations:

• From continuity: $$ 3 + \beta + \gamma = 6 \implies \beta + \gamma = 3 $$
• From first derivative continuity: $$ 2(3) + \beta = 5 \implies 6 + \beta = 5 \implies \beta = -1 $$

5. Finding $\gamma$

Substituting $\beta = -1$ into $\beta + \gamma = 3$ gives: $$ -1 + \gamma = 3 \implies \gamma = 4 $$