How do you find the absolute minimum and maximum on #[pi/2,pi/2]# of the function #f(x)=sinx^2#?
Hello,
Answer. Minimum is 0 and maximum is 1.
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To find the absolute minimum and maximum of the function ( f(x) = \sin(x^2) ) on the interval ([ \frac{\pi}{2}, \frac{\pi}{2}]), we first find the critical points by setting the derivative equal to zero and then evaluate the function at these points as well as at the endpoints of the interval.

Find the derivative of the function: [ f'(x) = 2x\cos(x^2) ]

Set the derivative equal to zero and solve for critical points: [ 2x\cos(x^2) = 0 ] [ x = 0 \text{ or } \cos(x^2) = 0 ]

Check endpoints of the interval: [ f(\frac{\pi}{2}) = \sin\left(\frac{\pi^2}{4}\right) ] [ f(\frac{\pi}{2}) = \sin\left(\frac{\pi^2}{4}\right) ]

Evaluate the function at the critical points: [ f(0) = \sin(0) = 0 ]

Evaluate the function at the critical points from (\cos(x^2) = 0): [ x = \pm \frac{\sqrt{2\pi}}{2} ] [ f\left(\pm \frac{\sqrt{2\pi}}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 ]

Compare all values to find the absolute minimum and maximum: [ \text{Absolute minimum: } 0 \text{ at } x = 0 ] [ \text{Absolute maximum: } 1 \text{ at } x = \pm \frac{\sqrt{2\pi}}{2} ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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