Calc2.47

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Find the absolute minimum and maximum on \left[-{\frac  {\pi }{2}},{\frac  {\pi }{2}}\right] of the function f(x)=\sin(x^{2}).

Again, any absolute minimum or maximum of this function on the indicated interval can only occur at a critical point of the function or at one of the endpoints of the interval.

f'(x)=(2x)\cos(x^{2})\,

This derivative is never undefined. It is equal to 0 at infinitely many values of x since the \cos u function is periodic but on our given interval, it is equal to 0 only at a finite number of points. This derivative is equal to 0 either when x=0 or when \cos(x^{2})=0.

The cosine function is equal to 0 at any point in the set \{...,-{\frac  {3\pi }{2}},-{\frac  {\pi }{2}},{\frac  {\pi }{2}},{\frac  {3\pi }{2}},...\}. So, the derivative will be equal to 0 whenever x^{2} is equal to one of these values. The only values of x inside of our interval for which this is true are -{\sqrt  {{\frac  {\pi }{2}}}} and {\sqrt  {{\frac  {\pi }{2}}}}. Thus, we have five values as candidates for our absolute max and the same five are candidates for our absolute min. Plug in all five values into the original function. The highest will be our absolute max and the lowest will be our absolute min.

f(-{\frac  {\pi }{2}})=\sin {\frac  {\pi ^{2}}{4}}\approx 0.624\,

f(-{\sqrt  {{\frac  {\pi }{2}}}})=1\,

f(0)=0\,

f({\sqrt  {{\frac  {\pi }{2}}}})=1\,

f({\frac  {\pi }{2}})=\sin {\frac  {\pi ^{2}}{4}}\approx 0.624\,

Thus, we have an absolute minimum on the interval at the point (0,0) and an absolute maximum on the interval at two points, both of which have the same height, \left(-{\sqrt  {{\frac  {\pi }{2}}}},1\right) and \left({\sqrt  {{\frac  {\pi }{2}}}},1\right).


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