6. Graphical Applications of Derivatives

The First Derivative Test

6. Graphical Applications of Derivatives

The First Derivative Test

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Multiple Choice

Identify the local minimum and maximum values of the given function, if any.
$f(t)=t^2\ln t$ , $t>0$

Local maximum of $1$ at $x=1$ , Local minimum of $-0.19$ at $x=\frac12$

Local maximum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local minima

Local minimum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local maxima

No local extrema

Verified step by step guidance

Step 1: To identify local extrema (local minimum or maximum), start by finding the critical points of the function. Critical points occur where the derivative of the function is zero or undefined. Begin by differentiating the given function f(t) = t^2 * ln(t). Use the product rule for differentiation: (uv)' = u'v + uv'.

Step 2: Apply the product rule to f(t). Let u = t^2 and v = ln(t). Compute u' = 2t and v' = 1/t. Substitute these into the product rule formula: f'(t) = u'v + uv' = 2t * ln(t) + t^2 * (1/t). Simplify the derivative to f'(t) = 2t * ln(t) + t.

Step 3: Set f'(t) = 0 to find the critical points. This gives the equation 2t * ln(t) + t = 0. Factor out t: t(2ln(t) + 1) = 0. Since t > 0 (as given in the problem), discard t = 0 as a solution. Solve 2ln(t) + 1 = 0 for t.

Step 4: Solve 2ln(t) + 1 = 0. Subtract 1 from both sides: 2ln(t) = -1. Divide by 2: ln(t) = -1/2. Exponentiate both sides to solve for t: t = e^(-1/2) = 1/sqrt(e). This is the critical point.

Step 5: Determine whether the critical point t = 1/sqrt(e) is a local minimum, maximum, or neither. Use the second derivative test or analyze the sign changes of f'(t) around the critical point. Compute f''(t) and evaluate its behavior at t = 1/sqrt(e). If f''(t) > 0, it is a local minimum; if f''(t) < 0, it is a local maximum. If f''(t) = 0, further analysis is needed.