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Let 
f(x)=(3x^(2)+1)/(x+2).
Find 
f^(')(-3).
Choose 1 answer:
(A) 
10
(B) -10
(C) -28
(D) 28

Let f(x)=3x2+1x+2 f(x)=\frac{3 x^{2}+1}{x+2} .\newlineFind f(3) f^{\prime}(-3) .\newlineChoose 11 answer:\newline(A) 10 \mathbf{1 0} \newline(B) 10-10\newline(C) 28-28\newline(D) 2828

Full solution

Q. Let f(x)=3x2+1x+2 f(x)=\frac{3 x^{2}+1}{x+2} .\newlineFind f(3) f^{\prime}(-3) .\newlineChoose 11 answer:\newline(A) 10 \mathbf{1 0} \newline(B) 10-10\newline(C) 28-28\newline(D) 2828
  1. Apply Quotient Rule: To find the derivative of the function f(x)=3x2+1x+2f(x) = \frac{3x^2 + 1}{x + 2}, we will use the quotient rule. The quotient rule states that if we have a function that is the quotient of two functions, u(x)v(x)\frac{u(x)}{v(x)}, then its derivative is given by v(x)u(x)u(x)v(x)(v(x))2\frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}. Here, u(x)=3x2+1u(x) = 3x^2 + 1 and v(x)=x+2v(x) = x + 2.
  2. Find u(x)u'(x): First, we need to find the derivative of u(x)=3x2+1u(x) = 3x^2 + 1. The derivative of 3x23x^2 is 6x6x, and the derivative of a constant is 00. Therefore, u(x)=6xu'(x) = 6x.
  3. Find v(x)v'(x): Next, we need to find the derivative of v(x)=x+2v(x) = x + 2. The derivative of xx is 11, and the derivative of a constant is 00. Therefore, v(x)=1v'(x) = 1.
  4. Apply Quotient Rule: Now we apply the quotient rule. The derivative of f(x)f(x), denoted as f(x)f'(x), is given by:\newlinef(x)=(x+2)(6x)(3x2+1)(1)(x+2)2.f'(x) = \frac{(x + 2) \cdot (6x) - (3x^2 + 1) \cdot (1)}{(x + 2)^2}.
  5. Simplify f(x)f'(x): We simplify the expression for f(x)f'(x):f(x)=6x2+12x3x21(x+2)2f'(x) = \frac{6x^2 + 12x - 3x^2 - 1}{(x + 2)^2}.
  6. Combine Like Terms: Further simplifying, we combine like terms in the numerator: f(x)=3x2+12x1(x+2)2f'(x) = \frac{3x^2 + 12x - 1}{(x + 2)^2}.
  7. Evaluate at x=3x = -3: Now we need to evaluate f(x)f'(x) at x=3x = -3. We substitute 3-3 into the derivative: f(3)=3(3)2+12(3)1((3)+2)2.f'(-3) = \frac{3(-3)^2 + 12(-3) - 1}{((-3) + 2)^2}.
  8. Calculate Numerator and Denominator: We calculate the numerator and the denominator separately:\newlineNumerator: 3(3)2+12(3)1=3(9)361=27361=103(-3)^2 + 12(-3) - 1 = 3(9) - 36 - 1 = 27 - 36 - 1 = -10.\newlineDenominator: ((3)+2)2=(1)2=1((-3) + 2)^2 = (-1)^2 = 1.
  9. Divide Numerator by Denominator: Now we divide the numerator by the denominator: f(3)=(10)/1=10f'(-3) = (-10) / 1 = -10.

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