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A curve has equation y=8 x (2 x-1)^-1. find the values of ( x ) at which the curve has a stationary point and determine the nature of each stationary point, justifying your answers.

a) The curve has a maximum at x=0.5x=0.5 and a minimum at x=0x=0.
b) The curve has a minimum at x=0.5x=0.5 and a maximum at x=0x=0.
c) The curve has a maximum at x=1x=1 and a minimum at x=0x=0.
d) The curve has a minimum at x=1x=1 and a maximum at x=0x=0.

User Loubna
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Final answer:

To find the values of x at which the curve has a stationary point, we need to solve for the values of x when the derivative of the function is equal to zero. The curve has a minimum at x=0 and a maximum at x=0.5.

Step-by-step explanation:

The curve has a stationary point where the derivative of the function is equal to zero.

To find these points, we need to find the derivative of the function y = 8x(2x-1)^-1.

Taking the derivative using the power rule and chain rule, we get: dy/dx = -16x2 + 8x. Setting this derivative equal to zero, we can solve for x to find the x-values of the stationary points.

Setting -16x2 + 8x = 0, we can factor out an x to get x(-16x + 8) = 0. So x = 0 or -16x + 8 = 0. Solving the second equation, we find x = 0.5.

To determine the nature of each stationary point, we can look at the second derivative.

Taking the derivative of -16x2 + 8x, we get: d2y/dx2 = -32x + 8. Evaluating the second derivative at the x-values of the stationary points, we find that d2y/dx2(0) = 8 and d2y/dx2(0.5) = -16.

Since the second derivative is positive at x = 0 and negative at x = 0.5, the curve has a minimum at x = 0 and a maximum at x = 0.5.

User Darko Miletic
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