Examine the derivability of the following functions at the specified points :
`x^(3)" at "x = 2`

Answer

To examine the derivability of the function \( f(x) = x^3 \) at the point \( x = 2 \), we will follow these steps:

### Step 1: Check Continuity of the Function
A function is continuous at a point if:
1. \( f(a) \) is defined.
2. \( \lim_{x \to a} f(x) \) exists.
3. \( \lim_{x \to a} f(x) = f(a) \).

For \( f(x) = x^3 \) at \( x = 2 \):
- \( f(2) = 2^3 = 8 \).
- We need to check the limit:
\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} x^3 = 2^3 = 8. \]
- Since \( f(2) = 8 \) and \( \lim_{x \to 2} f(x) = 8 \), the function is continuous at \( x = 2 \).

### Step 2: Find the Derivative of the Function
To check for differentiability, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3) = 3x^2. \]

### Step 3: Check the Continuity of the Derivative
Next, we need to check if \( f'(x) \) is continuous at \( x = 2 \):
- The derivative \( f'(x) = 3x^2 \) is a polynomial, and polynomials are continuous everywhere.
- Therefore, \( f'(x) \) is continuous at \( x = 2 \).

### Step 4: Calculate the Left-Hand and Right-Hand Derivatives
To confirm differentiability, we need to check the left-hand derivative and right-hand derivative at \( x = 2 \):
- Left-hand derivative:
\[ f'(2^-) = \lim_{h \to 0^-} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0^-} \frac{(2 + h)^3 - 8}{h}. \]
Expanding \( (2 + h)^3 \):
\[ (2 + h)^3 = 8 + 12h + 6h^2 + h^3. \]
Thus,
\[ f'(2^-) = \lim_{h \to 0^-} \frac{(8 + 12h + 6h^2 + h^3) - 8}{h} = \lim_{h \to 0^-} \frac{12h + 6h^2 + h^3}{h} = \lim_{h \to 0^-} (12 + 6h + h^2) = 12. \]

- Right-hand derivative:
\[ f'(2^+) = \lim_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0^+} \frac{(2 + h)^3 - 8}{h} = 12. \]

### Step 5: Conclusion
Since both the left-hand and right-hand derivatives at \( x = 2 \) are equal:
\[ f'(2^-) = f'(2^+) = 12, \]
the function \( f(x) = x^3 \) is differentiable at \( x = 2 \).

### Final Answer
The function \( f(x) = x^3 \) is derivable at \( x = 2 \).

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