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.
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Find the derivative of the function: [ f'(x) = 2x\cos(x^2) ]
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Set the derivative equal to zero and solve for critical points: [ 2x\cos(x^2) = 0 ] [ x = 0 \text{ or } \cos(x^2) = 0 ]
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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) ]
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Evaluate the function at the critical points: [ f(0) = \sin(0) = 0 ]
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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 ]
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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 one-sided 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 one-sided 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 one-sided 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 one-sided 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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