91Ó°ÊÓ

After understanding and evaluating the endpoints of a function's interval, the next step is to identify critical points. These are important because they can reveal where a function changes direction, potentially indicating maximum or minimum values. For the function \( h(t) = \sin(t^2) \), we find the critical points by taking its derivative. The derivative, \( h'(t) = 2t \cos(t^2) \), provides the slope at any point along the function. Critical points occur where the derivative is zero or undefined.

Setting \( 2t \cos(t^2) = 0 \) yields two conditions:

• \( 2t = 0 \Rightarrow t = 0 \): Already evaluated as an endpoint.
• \( \cos(t^2) = 0 \): To solve this, we look for where \( \cos(x) = 0 \), which occurs at \( x = (2k+1)\frac{\pi}{2} \) for integer \(k\).

Only values that lie within the specified interval \([0, \pi]\) are relevant. This ensures that the critical points we consider are indeed part of our function’s domain.

Trigonometric functions are essential in understanding periodic behavior, and \( h(t) = \sin(t^2) \) incorporates this by squaring the input to the sine function. This creates a unique curve because \( \sin(t^2) \) uses the square of \( t \) as its argument, rather than \( t \) itself, emphasizing behavior that's not simply repetitive over ordinary intervals like \([0, 2\pi]\).

Here are some characteristics of trigonometric functions like sine:

• Periodicity: While \( \sin(t) \) has a straightforward periodic interval of \([0, 2\pi]\), \( \sin(t^2) \) modifies this by increasing the rate of change within the function.
• Range: Sine values always oscillate between -1 and 1, no matter the input.
• Symmetry: Typically symmetrical about the origin for the simple \( \sin(t) \), this symmetry translates into varied oscillating patterns with \( \sin(t^2) \).

Understanding these behaviors helps predict the peaks and troughs within a given interval.

With the function \( h(t) = \sin(t^2) \) evaluated on the interval \([0, \pi]\), it's crucial to assess both the endpoints and any critical points within to determine global maximum and minimum values.

Endpoint Evaluation: Start by calculating the value of \( h(t) \) at the endpoints of the interval. For this problem, these are:

• At \( t = 0 \), \( h(0) = \sin(0^2) = 0 \).
• At \( t = \pi \), \( h(\pi) = \sin(\pi^2) \approx 0.4303 \).

Critical Point Evaluation: Critical points are determined by solving \( 2t \cos(t^2) = 0 \). Within the interval, points like \( t = \sqrt{\frac{\pi}{2}} \) yield specific function values, providing information about potential maxima or minima.

• At \( t = \sqrt{\frac{\pi}{2}} \), \( h(t) = \sin(\frac{\pi}{2}) = 1 \).
• At \( t = \sqrt{\frac{3\pi}{2}} \), \( h(t) = \sin(\frac{3\pi}{2}) = -1 \).

Finally, compare all these evaluated points to conclude the global maximum and minimum values within the interval, ensuring that none are overlooked.