let f(x) = x² / 4(x + 1). find all critical points of f. as your answer, please input the sum of all…

Explanation:

Step1: Find the derivative of (f(x))

Use the quotient - rule. If (f(x)=\frac{u(x)}{v(x)}), where (u(x) = x^{2}) and (v(x)=4(x + 1)), then (u'(x)=2x) and (v'(x)=4). The quotient - rule states that (f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^{2}}). So (f'(x)=\frac{2x\cdot4(x + 1)-x^{2}\cdot4}{16(x + 1)^{2}}=\frac{8x(x + 1)-4x^{2}}{16(x + 1)^{2}}=\frac{8x^{2}+8x-4x^{2}}{16(x + 1)^{2}}=\frac{4x^{2}+8x}{16(x + 1)^{2}}=\frac{x^{2}+2x}{4(x + 1)^{2}}).

Step2: Set the derivative equal to zero

Set (f'(x)=0), so (\frac{x^{2}+2x}{4(x + 1)^{2}} = 0). A fraction is zero when the numerator is zero and the denominator is non - zero. Set (x^{2}+2x=0), factor out an (x): (x(x + 2)=0). This gives (x = 0) or (x=-2). The denominator (4(x + 1)^{2}\neq0) when (x = 0) and (x=-2).

Step3: Check for points where the derivative is undefined

The derivative (f'(x)=\frac{x^{2}+2x}{4(x + 1)^{2}}) is undefined when (4(x + 1)^{2}=0), i.e., when (x=-1). But (x=-1) is not in the domain of (f(x)) (since (f(x)=\frac{x^{2}}{4(x + 1)}) and division by zero is not allowed), so it is not a critical point.

Step4: Calculate the sum of critical points

The critical points are (x = 0) and (x=-2). The sum of the critical points is (0+( - 2)=-2).

Answer:

(-2)