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a) f'(x) = 3x^2 -a =0

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a)F'(x)=3x^2-a=0
x=+-root a/3

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(a/3)^0.5

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if you find derivative of function and solve using quadratic formula you get x = sqrt(3a)/3 which means that there should be 2 different critical pts

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If a is positive

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If a is zero

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Trying to find Dr/dt(speed of the beam).

when the point is 3km, the beam is moving at 4 revolutions/min is given and we are trying to find the speed of the beam at when the point is 1km away

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If a is negative

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b) value of b do not matter, a<0, no critical point

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if a is zero; b doesn't matter; neither

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We want to find the instantaneous rate of change 1km from point p, we know

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c)value of b do not matter, a=0, one critical point

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a < 0, b doesn't matter

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Highlight variables and important key words such as decreasing, increasing, speed, etc.

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d) value of b do not matter, a>0, two critical points, the value of critical point change as value of a change.

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b) for a < 0 and b of all values,  f(x) has no critical point.

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c) for a = 0 and b of all values, f(x) has one critical point and that is x = 0. This is neither a local maximum or local minimum.

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Critical points:
x = + or - (a/3)^0.5

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d) for a > 0 and b of all values, f(x) has two critical points and that is x equals negative or positive root a/3. Point x = - (a/3)^1/2 gives the local maximum, while point x = (a/3) ^1 /2 gives the local minimum.

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a > 0, b doesn't matter