Question

Sports car racing is a form of motorsport which uses sports car prototypes. The competition is held on special tracks designed in various shapes. The equation of one such track is given as 

(i) Find \(f'(x)\) for \(0<x>3\). 
(ii) Find \(f'(4)\). 
(iii)(a) Test for continuity of \(f(x)\) at \(x=3\). 
OR 
(iii)(b) Test for differentiability of \(f(x)\) at \(x=3\). 
 

Hint:

For piecewise functions, always check continuity first using LHL, RHL and function value. If the function is continuous, then compare left and right derivatives to test differentiability.

Answer 1

(i) Find \( f'(x) \) for \( 0 \leq x < 3 \): 
Given the function for \( 0 \leq x < 3 \): \[ f(x) = x^4 - 4x^2 + 4 \] To find \( f'(x) \), we differentiate the function using standard differentiation rules: \[ f'(x) = \frac{d}{dx}(x^4 - 4x^2 + 4) \] Using the power rule: \[ f'(x) = 4x^3 - 8x \] Thus, the derivative is: \[ f'(x) = 4x^3 - 8x \] for \( 0 \leq x < 3 \). 

(ii) Find \( f'(4) \): 
Given the second part of the function for \( x \geq 3 \): \[ f(x) = x^2 + 40 \] To find \( f'(x) \), we differentiate: \[ f'(x) = \frac{d}{dx}(x^2 + 40) = 2x \] Now, evaluate at \( x = 4 \): \[ f'(4) = 2(4) = 8 \] Thus, \( f'(4) = 8 \). 

(iii)(a) Test for continuity of \( f(x) \) at \( x = 3 \): 
For \( f(x) \) to be continuous at \( x = 3 \), the left-hand limit, right-hand limit, and the value of the function at \( x = 3 \) must be equal. We have: - For \( 0 \leq x < 3 \), \( f(x) = x^4 - 4x^2 + 4 \), so: \[ \lim_{x \to 3^-} f(x) = 3^4 - 4(3^2) + 4 = 81 - 36 + 4 = 49 \] - For \( x \geq 3 \), \( f(x) = x^2 + 40 \), so: \[ \lim_{x \to 3^+} f(x) = 3^2 + 40 = 9 + 40 = 49 \] - The value of \( f(x) \) at \( x = 3 \) is also: \[ f(3) = 3^2 + 40 = 9 + 40 = 49 \] Since the left-hand limit, right-hand limit, and the value of the function are all equal, \( f(x) \) is continuous at \( x = 3 \). 

(iii)(b) Test for differentiability of \( f(x) \) at \( x = 3 \): 
For \( f(x) \) to be differentiable at \( x = 3 \), the left-hand derivative and right-hand derivative must be equal. We have: - For \( 0 \leq x < 3 \), \( f'(x) = 4x^3 - 8x \), so: \[ \lim_{x \to 3^-} f'(x) = 4(3)^3 - 8(3) = 4(27) - 24 = 108 - 24 = 84 \] - For \( x \geq 3 \), \( f'(x) = 2x \), so: \[ \lim_{x \to 3^+} f'(x) = 2(3) = 6 \] Since the left-hand derivative and right-hand derivative are not equal, \( f(x) \) is not differentiable at \( x = 3 \). 

Final Answers: 
1. \( f'(x) = 4x^3 - 8x \) for \( 0 \leq x < 3 \). 
2. \( f'(4) = 8 \). 
3. \( f(x) \) is continuous at \( x = 3 \). 
4. \( f(x) \) is not differentiable at \( x = 3 \).