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The curve with equation y = x^2 - 32 adix{(x)} + 20, x > 0 has a stationary point P - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 4
Question 8
The curve with equation y = x^2 - 32 adix{(x)} + 20, x > 0 has a stationary point P. Use calculus a) to find the coordinates of P, b) to determine the nature... show full transcript
Worked Solution & Example Answer:The curve with equation y = x^2 - 32 adix{(x)} + 20, x > 0 has a stationary point P - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 4
Step 1
to find the coordinates of P,
Answer
To find the coordinates of the stationary point P, we first need to differentiate the given function with respect to x:
adix{x}} = 2x - rac{16}{ adix{x}}$$ Setting the derivative equal to zero gives: $$2x - rac{16}{ adix{x}} = 0$$ Rearranging, we have: $$2x = rac{16}{ adix{x}}$$ Multiplying both sides by adix{x} results in: $$2x adix{x} = 16$$ To eliminate the square root, we square both sides: $$4x^3 = 256$$ Dividing both sides by 4 yields: $$x^3 = 64$$ Taking the cube root of both sides gives: $$x = 4$$ Now substituting this value of x back into the original equation to find y: $$y = (4)^2 - 32 adix{(4)} + 20$$ This simplifies to: $$y = 16 - 32(2) + 20 = 16 - 64 + 20 = -28$$ Thus, the coordinates of P are (4, -28).Step 2
to determine the nature of the stationary point P,
Answer
To determine the nature of the stationary point P, we need to evaluate the second derivative of the function:
First, we compute the first derivative again:
adix{x}}$$ Now, to find the second derivative: $$ rac{d^2y}{dx^2} = 2 + rac{8}{x^{3/2}}$$ Next, we evaluate the second derivative at the stationary point x = 4: $$ rac{d^2y}{dx^2} igg|_{x=4} = 2 + rac{8}{(4)^{3/2}} = 2 + rac{8}{8} = 2 + 1 = 3$$ Since the second derivative at point P is positive (3 > 0), it indicates that the stationary point is a minimum.