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A critical point is where the derivative of a function is zero or undefined. Finding these points can help determine where a function's graph might change direction. To find critical points, you first take the derivative of the function. In the case of the function \( h(\theta) = 3 \theta^{2/3} \), the process of taking the derivative resulted in \( h'(\theta) = 2 \theta^{-1/3} \).
Interestingly, when setting the derivative to zero, \( h'(\theta) = 0 \), the equation has no solution since a power of a positive constant never equals zero. However, \( h'(\theta) \) is undefined at \( \theta = 0 \) because you cannot raise zero to a negative power. This is an example where even if a derivative doesn't equate to zero, a critical point can still exist due to being undefined. Hence, \( \theta = 0 \) is a critical point.

The derivative of a function provides a powerful tool in calculus. It represents the rate at which a function is changing at any point. Essentially, it's like a radar showing how your function increases or decreases as you move along the curve.
To find the derivative of \( h(\theta) = 3 \theta^{2/3} \), we use the power rule, which says to multiply the exponent by the coefficient and decrease the exponent by one.For this function, let’s go through the steps:

• The exponent \( 2/3 \) is multiplied by the coefficient 3, giving \( 2 \).
• The new exponent becomes \( 2/3 - 1 = -1/3 \).

Hence, the derivative is \( h'(\theta) = 2 \theta^{-1/3} \). With derivatives, negative exponents can indicate points where the function or its derivative is undefined, which is handy when checking for critical points.

When we talk about absolute maximum and minimum, we're referring to the highest and lowest values a function attains over a given interval. These points are critical for understanding overall behavior.
To find these values for \( h(\theta) = 3 \theta^{2/3} \) in the interval \(-27 \leq \theta \leq 8\), we check both the endpoints and any critical points. By calculating:

• At \( \theta = -27 \), \( h(-27) = 27 \).
• At \( \theta = 8 \), \( h(8) = 12 \).
• At the critical point \( \theta = 0 \), \( h(0) = 0 \).

The absolute maximum value here is 27 at \( \theta = -27 \), and the absolute minimum value is 0 at \( \theta = 0 \). This shows how checking both endpoints and critical points together ensures you capture the full range of the function's values within the interval.

Interval evaluation is crucial when determining the absolute maximum and minimum values of a function. It involves observing how a function behaves at specified boundary points known as endpoints, as well as at any critical points identified within the interval.
For the function \( h(\theta) = 3 \theta^{2/3} \) on the interval \(-27 \leq \theta \leq 8\), evaluating the function means plugging the endpoint values and any critical point values into the original function:

• Evaluate at endpoints to get immediate boundary values.
• Include critical point evaluations — key locations within the interval that may represent extreme values.

Through interval evaluation, we firmly establish that the highest value \( 27 \) occurs at one endpoint (\( \theta = -27 \)) and the lowest \( 0 \) at the critical point (\( \theta = 0 \)), providing a complete picture of the function's behavior across the given interval.