▶️ Answer/Explanation
Solution
For \( f(x) = x^6 \), the derivative is found using the power rule: \( f'(x) = 6x^{6-1} = 6x^5 \).
✅ Answer: C)
▶️ Answer/Explanation
Solution
For \( f(x) = \frac{1}{x^6} = x^{-6} \), the derivative is found using the power rule: \( f'(x) = -6x^{-6-1} = -6x^{-7} = -\frac{6}{x^7} \).
✅ Answer: D)
▶️ Answer/Explanation
Solution
For \( f(x) = \sqrt[3]{x} = x^{1/3} \), the derivative is found using the power rule: \( f'(x) = \frac{1}{3} x^{1/3 – 1} = \frac{1}{3} x^{-2/3} \).
✅ Answer: C)
▶️ Answer/Explanation
Solution
For the line \( y = -\frac{1}{3}x + b \) to be normal to the curve \( y = x^3 \), their slopes must be negative reciprocals at the point of intersection.
The derivative of \( y = x^3 \) is \( \frac{dy}{dx} = 3x^2 \). At the point of tangency, the slope of the curve is \( 3x^2 \), and the slope of the line is \( -\frac{1}{3} \).
Set \( 3x^2 = -\frac{1}{-\frac{1}{3}} = 3 \), so \( 3x^2 = 3 \), \( x^2 = 1 \), \( x = \pm 1 \). Since \( b \) is non-negative, consider the positive \( x = 1 \).
At \( x = 1 \), \( y = 1^3 = 1 \). The line equation is \( y = -\frac{1}{3}x + b \), so \( 1 = -\frac{1}{3}(1) + b \), \( 1 = -\frac{1}{3} + b \), \( b = 1 + \frac{1}{3} = \frac{4}{3} \).
Check \( x = -1 \): \( y = (-1)^3 = -1 \), \( -1 = -\frac{1}{3}(-1) + b \), \( -1 = \frac{1}{3} + b \), \( b = -1 – \frac{1}{3} = -\frac{4}{3} \) (negative, discarded).
Thus, \( b = \frac{4}{3} \).
✅ Answer: C)