Question

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.

Answer 1

Answer:

The curve has a stationary point at (a,14), where a is a positive constant.the value of a is 46788i

Answer 2

Answer:

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