Question

Find any critical numbers of the function. $$ g(t)=t \sqrt{4-t}, t<3 $$

Short answer

The critical number for the function \(g(t) = t \cdot \sqrt{4 - t}\) for \(t < 3\) is \(t = \frac{8}{3}\).

Step-by-step solution

Applying the Product Rule for Derivatives.

First differentiate the given function. Since g(t) is a product of two functions, we will apply the product rule of derivates which states: if a function \(h(t) = f(t) \cdot g(t)\), then its derivative is given by \(h'(t) = f'(t) \cdot g(t) + f(t) \cdot g'(t)\). Applying this rule to the given function \(g(t) = t \cdot \sqrt{4-t}\), we obtain:\(g'(t) = 1 \cdot \sqrt{4-t} + t \cdot \frac{1}{2\sqrt{4-t}}(-1)\)

Simplifying the first derivative and finding critical points.

Simplify the derivative as: \(g'(t) = \sqrt{4-t} - \frac{t}{2 \sqrt{4-t}}\). The critical numbers are those \(t\) values that make \(g'(t) = 0\) or the values that make \(g'(t)\) undefined. Solve the equation \(g'(t) = 0\), by first multiplying both sides by \(2 \sqrt{4-t}\) to get: \(0 = 2(4-t) - t = 8 - 3t\). Solving this for \(t\), we get: \(t = \frac{8}{3}\). The derivative is undefined at \(t=4\) but because this value is outside the given interval (t<3), we disregard it. So, \(t = \frac{8}{3}\) is the only critical number exists on our interval.

Verifying the validity of the critical number.

Verify whether the derived critical number falls within the allocated interval. Since \(t = \frac{8}{3}\) is less than 3, it is a valid critical number for the given function. Thus, we conclude that the function \(g(t) = t \cdot \sqrt{4 - t}\) for \(t < 3\) has one critical number at \(t = \frac{8}{3}\).