▶️ Answer/Explanation
Solution
(a) Derivative of \( f \):
\[ f(x) = (25 – x^{2})^{1/2} \] \[ f'(x) = \frac{1}{2}(25 – x^{2})^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{25 – x^{2}}} \]
(b) Tangent Line at \( x = -3 \):
\[ f(-3) = \sqrt{25 – 9} = 4 \] \[ f'(-3) = \frac{3}{4} \]
Equation: \( y – 4 = \frac{3}{4}(x + 3) \) or \( y = \frac{3}{4}x + \frac{25}{4} \)
(c) Continuity of \( g \) at \( x = -3 \):
Check three conditions:
- \( g(-3) = f(-3) = 4 \) (defined)
- \( \lim_{x \to -3^-} g(x) = f(-3) = 4 \)
- \( \lim_{x \to -3^+} g(x) = -3 + 7 = 4 \)
Since all three conditions are satisfied and equal, \( g \) is continuous at \( x = -3 \).
(d) Integral Evaluation:
Let \( u = 25 – x^2 \), \( du = -2x dx \):
\[ \int_{0}^{5} x \sqrt{25 – x^{2}} \, dx = -\frac{1}{2} \int_{25}^{0} u^{1/2} \, du = \frac{1}{2} \int_{0}^{25} u^{1/2} \, du \] \[ = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_0^{25} = \frac{1}{3} (125 – 0) = \frac{125}{3} \]