▶️ Answer/Explanation
Solution
1. Rewrite the curve equation: \( y = -\frac{x^2}{2} \) (a downward parabola)
2. Distance squared to point \(\left(0,-\frac{1}{2}\right)\):
\( D^2 = (x-0)^2 + \left(-\frac{x^2}{2} – (-\frac{1}{2})\right)^2 = x^2 + \left(\frac{1 – x^2}{2}\right)^2 \)
3. Simplify distance function:
\( D^2 = x^2 + \frac{(1 – x^2)^2}{4} = \frac{x^4 – 2x^2 + 1 + 4x^2}{4} = \frac{x^4 + 2x^2 + 1}{4} \)
4. Find critical points by taking derivative:
\( \frac{dD^2}{dx} = \frac{4x^3 + 4x}{4} = x^3 + x \)
Set equal to zero: \( x(x^2 + 1) = 0 \) ⇒ \( x = 0 \) (only real solution)
5. Verify minimum:
Second derivative: \( \frac{d^2D^2}{dx^2} = 3x^2 + 1 > 0 \) for all x ⇒ minimum at x=0
6. Find corresponding y-value:
When \( x = 0 \), \( y = -\frac{0^2}{2} = 0 \)
✅ Answer: B) 0
▶️ Answer/Explanation
Solution
1. Given function: \( h(x) = (2f(x)+3)(1+g(x)) \)
2. Apply product rule for differentiation:
\( h'(x) = (2f(x)+3)’ \cdot (1+g(x)) + (2f(x)+3) \cdot (1+g(x))’ \)
\( h'(x) = (2f'(x))(1+g(x)) + (2f(x)+3)(g'(x)) \)
3. Evaluate at x = 1 using table values:
\( f(1) = 3 \), \( f'(1) = -2 \), \( g(1) = -3 \), \( g'(1) = 4 \)
4. Calculate each component:
\( 2f(1)+3 = 2(3)+3 = 9 \)
\( 1+g(1) = 1+(-3) = -2 \)
\( 2f'(1) = 2(-2) = -4 \)
5. Plug into derivative formula:
\( h'(1) = (-4)(-2) + (9)(4) = 8 + 36 = 44 \)
✅ Answer: D) 44