Try \( C(t) = \frac{10}{(t/2 + 1)} = \frac{20}{t+2} \)? Then \( C' = -20/(t+2)^2 \). Set = -5: \( (t+2)^2 = 4 \Rightarrow t = 2 \).
["Understanding the Function ( C(t) = \frac{10}{(t/2 + 1)} = \frac{20}{t+2} ) and Its Derivative – Finding Critical Points", "In calculus and applied mathematics, analyzing functions involves more than just evaluating their outputs. Understanding their derivatives is crucial for determining rates of change, optimization, and solving real-world problems. One classic example is the function:", "[\nC(t) = \frac{10}{(t/2 + 1)} = \frac{20}{t + 2}\n]", "This article explores how to derive this function, compute its derivative, and solve for key points such as when ( C'(t) = -5 )—a common type of problem in applied calculus and science.", "---", "### Simplifying the Function", "The original expression is:", "[\nC(t) = \frac{10}{\frac{t}{2} + 1}\n]", "Simplify the denominator:", "[\n\frac{t}{2} + 1 = \frac{t + 2}{2}\n]", "So,", "[\nC(t) = \frac{10}{\frac{t + 2}{2}} = 10 \cdot \frac{2}{t + 2} = \frac{20}{t + 2}\n]", "This confirms:", "[\nC(t) = \frac{20}{t + 2}\n]", "---", "### Computing the Derivative ( C'(t) )", "Using the quotient rule or recognizing ( C(t) = 20(t+2)^{-1} ), we apply the chain rule:", "[\nC'(t) = 20 \cdot (-1)(t + 2)^{-2} \cdot (1) = -\frac{20}{(t + 2)^2}\n]", "Thus,", "[\nC'(t) = -\frac{20}{(t + 2)^2}\n]", "---", "### Solving for When the Derivative Equals -5", "We now solve for ( t ) when ( C'(t) = -5 ):", "[\n-\frac{20}{(t + 2)^2} = -5\n]", "Multiply both sides by ( -1 ):", "[\n\frac{20}{(t + 2)^2} = 5\n]", "Multiply both sides by ( (t + 2)^2 ):", "[\n20 = 5(t + 2)^2\n]", "Divide both sides by 5:", "[\n4 = (t + 2)^2\n]", "Take the square root of both sides:", "[\nt + 2 = \pm 2\n]", "Solve for ( t ):", "- ( t + 2 = 2 \Rightarrow t = 0 )\n- ( t + 2 = -2 \Rightarrow t = -4 )", "However, note that the original function ( C(t) = \frac{20}{t + 2} ) is undefined for ( t = -2 ). But values beyond that, including ( t = -4 ), are valid.", "---", "### Final Answer", "The values of ( t ) where the derivative ( C'(t) = -5 ) are:", "[\n\boxed{t = 0} \quad \ ext{and} \quad \boxed{t = -4}\n]", "---", "### Practical Insight", "This Example illustrates how derivatives of rational functions model real-world decay or growth processes. Finding when the slope equals a specific value helps identify inflection behavior—useful in physics, economics, and engineering. For ( C(t) ), knowing ( C'(t) = -5 ) identifies points where the quantity decreases at a constant rate of 5 units per time interval.", "---", "### Key Takeaways", "- Simplify rational functions for easier differentiation.\n- Derivative ( C'(t) = -\frac{20}{(t + 2)^2} ) always negative (since denominator > 0), modeling a decreasing function.\n- Solving ( C'(t) = -5 ) finds critical slowing points or rates in applied models.\n- Domain considerations prevent division by zero but allow meaningful solutions like ( t = -4 ).", "---", "Use keywords:\nTry ( C(t) = \frac{10}{(t/2 + 1)} ), derivative ( C'(t) = -20/(t+2)^2 ), solve ( C'(t) = -5 ), critical points, calculus applications, rational function derivatives.", "---", "Understand calculus concepts deeply—such as derivatives of transformations of simple functions—and apply them confidently to solve equations and model real-life systems."]
