What choices of α, β, and γ does the function f(x) = { x^3, x ≤ 1; αx^2 + βx + γ, x > 1 } have for being twice differentiable at x = 1?

Steps to Solve

1. Set up the piecewise function and continuity requirements

The function is defined as:

$$ f(x) = \begin{cases} x^3, & x \leq 1 \ \alpha x^2 + \beta x + \gamma, & x > 1 \end{cases} $$

We want to ensure that $f(1)$ and its derivatives are continuous at $x = 1$.

2. Ensure continuity of the function at x = 1

Set $f(1)$ from both parts equal to each other:

$$ 1^3 = \alpha(1^2) + \beta(1) + \gamma $$

which simplifies to:

$$ 1 = \alpha + \beta + \gamma \tag{1} $$

3. Ensure continuity of the first derivative at x = 1

Calculate the derivatives:

$$ f'(x) = \begin{cases} 3x^2, & x < 1 \ 2\alpha x + \beta, & x > 1 \end{cases} $$

Evaluate the derivatives at $x = 1$:

$$ f'(1^-) = 3(1^2) = 3 $$

Set the two derivatives equal:

$$ 3 = 2\alpha(1) + \beta \tag{2} $$

4. Ensure continuity of the second derivative at x = 1

Calculate the second derivatives:

$$ f''(x) = \begin{cases} 6x, & x < 1 \ 2\alpha, & x > 1 \end{cases} $$

Evaluate at $x = 1$:

$$ f''(1^-) = 6(1) = 6 $$

Set them equal:

$$ 6 = 2\alpha \tag{3} $$

5. Solve the equations simultaneously

From equation (3), solve for $\alpha$:

$$ \alpha = 3 $$

Substitute $\alpha$ into equation (2):

$$ 3 = 2(3) + \beta \implies 3 = 6 + \beta \implies \beta = -3 $$

Now substitute $\alpha$ and $\beta$ into equation (1):

$$ 1 = 3 - 3 + \gamma \implies \gamma = 1 $$

6. Final values for parameters

The values are:

• $\alpha = 3$
• $\beta = -3$
• $\gamma = 1$