1. Suppose that, for x ≥ 0, the curve C has equation y = 3x − x√x.(a) Find the coordinates of the x-intercepts of C.(4 marks)(b) Find the coordinates of the stationary point of C and determine whether it isa maximum, a minimum or a point of inflection.(7 marks)(c) Sketch the curve C.
Question
- Suppose that, for x ≥ 0, the curve C has equation y = 3x − x√x.(a) Find the coordinates of the x-intercepts of C.(4 marks)(b) Find the coordinates of the stationary point of C and determine whether it isa maximum, a minimum or a point of inflection.(7 marks)(c) Sketch the curve C.
Solution
(a) The x-intercepts of the curve C are the points where y = 0. So we set the equation 3x - x√x = 0 and solve for x.
This can be rewritten as x(3 - √x) = 0.
Setting each factor equal to zero gives the solutions x = 0 and x = 9. So, the x-intercepts of C are (0,0) and (9,0).
(b) To find the stationary points of C, we need to find where the derivative of y with respect to x is equal to zero.
The derivative of y = 3x - x√x is dy/dx = 3 - (1/2)√x - √x.
Setting this equal to zero gives the equation 3 - (1/2)√x - √x = 0.
Solving this equation gives x = 4.
Substituting x = 4 into the equation for C gives y = 3*4 - 4√4 = 8.
So, the stationary point of C is (4,8).
To determine whether this is a maximum, a minimum or a point of inflection, we need to find the second derivative of y with respect to x.
The second derivative of y = 3x - x√x is d²y/dx² = -(1/4)x^(-1/2) - 1.
Substituting x = 4 into this equation gives d²y/dx² = -(1/4)*4^(-1/2) - 1 = -1.5.
Since this is negative, the stationary point (4,8) is a maximum.
(c) To sketch the curve C, plot the points (0,0), (4,8), and (9,0).
The curve starts at the origin (0,0), rises to a maximum at (4,8), and then falls back to the x-axis at (9,0).
Between x = 0 and x = 4, the curve is increasing, and between x = 4 and x = 9, the curve is decreasing.
The curve is also concave down for all x ≥ 0, because the second derivative is negative.
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