The gradient of a curve at the point (x,y) is given by dy/dx=2(x+3)^1/2-x. The curve has a stationary point at (a,14), where a is a positive constant. Find the value of a.
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The curve has a stationary point at (a,14), where a is a positive constant.the value of a is 46788i
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Step-by-step explanation:
dy/dx = 0
subsitute a in x
0 = 2(a+3)^1/2 - a
a = 2(a+3)^1/2
take square on both sides
a^2 = [2(a+3)^1/2 ]^2
a^2 = 4(a+3)
a^2 = 4a + 12
a^2 - 4a - 12 = 0
a^2 - 6a + 2a - 12 = 0
a (a - 6) +2 (a - 6) +0
a - 6 = 0 , a + 2 = 0
a = 6 , a = - 2
a is a positive constant hence:
a = 6
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