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G(x) = 1/2 x^2sin2x+1/2 xcos2x - 1/4 sin2x

find the maximum and minimum values for G(x) on the interval [0,pi/2]

i found G'(x)= x^2 cos2x and i know that there is a global max at pi/4 but i can't find the min value for G(x)...

- Thread starter ronho1234
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G(x) = 1/2 x^2sin2x+1/2 xcos2x - 1/4 sin2x

find the maximum and minimum values for G(x) on the interval [0,pi/2]

i found G'(x)= x^2 cos2x and i know that there is a global max at pi/4 but i can't find the min value for G(x)...

- #2

Mentallic

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The max and min values can be found either at the turning points of the function, or on the boundaries. So from G'(x)=0 we can see we need to test x=0 and x=[itex]\pi[/itex]/4 but we also have the boundary conditions and it's quite possible that the min or max lies on the edges, so at x=0 and x=[itex]\pi[/itex]/2 (clearly x=0 overlaps so we aren't going to test it twice).i found G'(x)= x^2 cos2x and i know that there is a global max at pi/4 but i can't find the min value for G(x)...

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but since pi/2 is a boundary how do i know if its a max or min

do i substitute it back into g(x) and g'(x) if so i get -pi/4 and -1 respectively

sorry i still don't quite get it, could you please show me hot to work it out

- #4

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If I have some function y=f(x) then what are we given when we substitute some constant in for x, say, x=1?

What if we plug this x-value value into the derivative, y=f'(x)?

- #5

Ray Vickson

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If you want to maximize a smooth function f(x) on an interval [a,b] (that is, on the interval a ≤ x ≤ b that includes both endpoints), then: (i) in order that x=a be a (local) max it is *necessary* to have f'(a) ≤ 0 (that is, f(x) must _not be strictly increasing_ just to the right of a); (ii) in order tht x = b be a local max, it is necessary to have f'(b) ≥ 0 (so that f(x) is not strictly decreasing just to the left of b).

but since pi/2 is a boundary how do i know if its a max or min

do i substitute it back into g(x) and g'(x) if so i get -pi/4 and -1 respectively

sorry i still don't quite get it, could you please show me hot to work it out

Of course, the conditions for a min are the opposite of the above.

Neither of these conditions is *sufficient* unless the inequalities are strict; that is, you can have f'(a) = 0, and x = a can be either a max or a min.

RGV

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Or you could just as simply calculate the value of f(a) and f(b) to determine whether it's the min/max for that continuous interval.If you want to maximize a smooth function f(x) on an interval [a,b] (that is, on the interval a ≤ x ≤ b that includes both endpoints), then: (i) in order that x=a be a (local) max it is *necessary* to have f'(a) ≤ 0 (that is, f(x) must _not be strictly increasing_ just to the right of a); (ii) in order tht x = b be a local max, it is necessary to have f'(b) ≥ 0 (so that f(x) is not strictly decreasing just to the left of b).

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Ray Vickson

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OK, but the derivative will tell you whether nearby points are better or worse then the endpoint. Alternatively, you could compute f(x) at an endpoint and at another point very near the endpoint, to check how the function is behaving near the boundary.Or you could just as simply calculate the value of f(a) and f(b) to determine whether it's the min/max for that continuous interval.

RGV

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I'm not disagreeing with you that checking the derivative at the endpoints is usually a better method, I'm just offering the OP another (probably more simple) way of understanding how to find min/max values on an interval.OK, but the derivative will tell you whether nearby points are better or worse then the endpoint. Alternatively, you could compute f(x) at an endpoint and at another point very near the endpoint, to check how the function is behaving near the boundary.

RGV

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Ray Vickson

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Unfortunately, just computing f at the endpoints is not enough; we also need some information about the behaviour of f near the endpoints (is it increasing? decreasing?) One way, as I suggested, is to evaluate f'(a) and f'(b) (as well as f(a) and f(b)); another way is to reason it out: if y < b is the right-most stationary point in (a,b), then f will be decreasing between y and b if y is a max, and will be increasing between y and b if y is a min, etc. *Some* kind of information of that type is required.I'm not disagreeing with you that checking the derivative at the endpoints is usually a better method, I'm just offering the OP another (probably more simple) way of understanding how to find min/max values on an interval.

Anyway, testing f' at the endpoints is included as a vital part of most standard numerical optimization routines that utilize derivative methods.

RGV

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