▶️ Answer/Explanation
Solution
For \( f(x) = \frac{3x^2 – 1}{4x + 1} \), use the quotient rule: \( f'(x) = \frac{(4x + 1)(6x) – (3x^2 – 1)(4)}{(4x + 1)^2} \).
Numerator: \( (4x + 1)(6x) – (3x^2 – 1)(4) = 24x^2 + 6x – (12x^2 – 4) = 12x^2 + 6x + 4 \). So, \( f'(x) = \frac{12x^2 + 6x + 4}{(4x + 1)^2} \).
Evaluate at \( x = -1 \): Numerator = \( 12(-1)^2 + 6(-1) + 4 = 12 – 6 + 4 = 10 \); denominator = \( (4(-1) + 1)^2 = (-3)^2 = 9 \). Thus, \( f'(-1) = \frac{10}{9} \).
✅ Answer: C)
Question
The graphs of the functions f and g are shown above. If \( h(x)=\frac{f(x)+1}{g(x)+3x} \), then \( h'(2)= \)
A) \(\frac{1}{2}\)
B) \(\frac{9}{100}\)
C) \(\frac{1}{100}\)
D) \(\frac{1}{10}\)
▶️Answer/Explanation
Answer: C
Explanation:
1. Apply Quotient Rule:
\(h'(x) = \frac{(g(x)+3x)f'(x)-(f(x)+1)(g'(x)+3)}{(g(x)+3x)^2} \)
2. Evaluate at x=2:
From the graphs:
- \( f(2) = 3 \)
- \( f'(2) = \frac{1}{2} \)
- \( g(2) = 4 \)
- \( g'(2) = -2 \)
3. Substitute values:
\( h'(2) = \frac{(4+6)\left(\frac{1}{2}\right)-(3+1)(-2+3)}{(4+6)^2} \)
\(= \frac{10 \times 0.5 – 4 \times 1}{100} \)
\( = \frac{5 – 4}{100} = \frac{1}{100} \)
▶️ Answer/Explanation
Solution
To find the slope, compute the derivative \( y’ \) using the quotient rule. Let \( u = 4x^3 \), \( v = x + 3 \), so \( u’ = 12x^2 \), \( v’ = 1 \).
The quotient rule gives: \( y’ = \frac{u’v – uv’}{v^2} = \frac{(12x^2)(x + 3) – (4x^3)(1)}{(x + 3)^2} \).
Numerator: \( 12x^2 (x + 3) – 4x^3 = 12x^3 + 36x^2 – 4x^3 = 8x^3 + 36x^2 \). So, \( y’ = \frac{8x^3 + 36x^2}{(x + 3)^2} \).
Evaluate at \( x = 1 \): Numerator = \( 8(1)^3 + 36(1)^2 = 8 + 36 = 44 \); denominator = \( (1 + 3)^2 = 4^2 = 16 \). Thus, \( y'(1) = \frac{44}{16} = \frac{11}{4} \).
✅ Answer: B)
▶️ Answer/Explanation
Solution
For \( f(x) = \frac{x – 1}{x + 1} \) with \( x \neq -1 \), use the quotient rule. Let \( u = x – 1 \), \( v = x + 1 \), so \( u’ = 1 \), \( v’ = 1 \).
The quotient rule gives: \( f'(x) = \frac{u’v – uv’}{v^2} = \frac{(1)(x + 1) – (x – 1)(1)}{(x + 1)^2} \).
Numerator: \( (x + 1) – (x – 1) = x + 1 – x + 1 = 2 \). So, \( f'(x) = \frac{2}{(x + 1)^2} \).
Evaluate at \( x = 1 \): \( f'(1) = \frac{2}{(1 + 1)^2} = \frac{2}{2^2} = \frac{2}{4} = \frac{1}{2} \).
✅ Answer: D)