To determine the values of \( a \) and \( b \) that make the function
\[ f(x) = \begin{cases} ax^3 - 6x & \text{if } x < 1 \ bx^2 + 2 & \text{if } x > 1 \end{cases} \]
differentiable at \( x = 1 \), we need to ensure that \( f(x) \) is both continuous and has a continuous derivative at that point.
Step 1: Continuity at \( x = 1 \)
For \( f(x) \) to be continuous at \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]
Calculating \( \lim_{x \to 1^-} f(x) \):
\[ \lim_{x \to 1^-} f(x) = a(1)^3 - 6(1) = a - 6 \]
Calculating \( \lim_{x \to 1^+} f(x) \):
\[ \lim_{x \to 1^+} f(x) = b(1)^2 + 2 = b + 2 \]
Setting these two limits equal for continuity:
\[ a - 6 = b + 2 \]
Rearranging gives:
\[ a - b = 8 \quad \text{(1)} \]
Step 2: Differentiability at \( x = 1 \)
For \( f(x) \) to be differentiable at \( x = 1 \), the derivatives from the left and right must also be equal at that point.
Calculating the derivative for \( x < 1 \):
\[ f'(x) = \frac{d}{dx}(ax^3 - 6x) = 3ax^2 - 6 \]
Calculating the left-hand derivative at \( x = 1 \):
\[ \lim_{x \to 1^-} f'(x) = 3a(1)^2 - 6 = 3a - 6 \]
Calculating the derivative for \( x > 1 \):
\[ f'(x) = \frac{d}{dx}(bx^2 + 2) = 2bx \]
Calculating the right-hand derivative at \( x = 1 \):
\[ \lim_{x \to 1^+} f'(x) = 2b(1) = 2b \]
Setting the derivatives equal at \( x = 1 \):
\[ 3a - 6 = 2b \quad \text{(2)} \]
Step 3: Solve the system of equations
We now have a system of two equations:
- \( a - b = 8 \)
- \( 3a - 6 = 2b \)
From equation (1), we can express \( a \) in terms of \( b \):
\[ a = b + 8 \]
Substituting this into equation (2):
\[ 3(b + 8) - 6 = 2b \] \[ 3b + 24 - 6 = 2b \] \[ 3b + 18 = 2b \] \[ 3b - 2b = -18 \] \[ b = -18 \]
Now substituting \( b \) back to find \( a \):
\[ a = -18 + 8 = -10 \]
Final result
The values that make \( f(x) \) differentiable at \( x = 1 \) are:
\[ a = -10 \quad \text{and} \quad b = -18 \]