The curve C has equation
y = 3x^3 - 8x^2 - 3
(a) Find
(i) \( \frac{dy}{dx} \)
(ii) \( \frac{d^2y}{dx^2} \)
(b) Verify that C has a stationary point when \( x = 2 \)
(c) Determine the nature of this stationary point, giving a reason for your answer. - Edexcel - A-Level Maths Pure - Question 3 - 2017 - Paper 1
Question 3
The curve C has equation
y = 3x^3 - 8x^2 - 3
(a) Find
(i) \( \frac{dy}{dx} \)
(ii) \( \frac{d^2y}{dx^2} \)
(b) Verify that C has a stationary point when \( x ... show full transcript
Worked Solution & Example Answer:The curve C has equation
y = 3x^3 - 8x^2 - 3
(a) Find
(i) \( \frac{dy}{dx} \)
(ii) \( \frac{d^2y}{dx^2} \)
(b) Verify that C has a stationary point when \( x = 2 \)
(c) Determine the nature of this stationary point, giving a reason for your answer. - Edexcel - A-Level Maths Pure - Question 3 - 2017 - Paper 1
Step 1
Find (i) \( \frac{dy}{dx} \)
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Answer
To find ( \frac{dy}{dx} ), we differentiate the equation of the curve:
[ y = 3x^3 - 8x^2 - 3 ]
Applying the power rule:
[ \frac{dy}{dx} = 9x^2 - 16x ]
Step 2
Find (ii) \( \frac{d^2y}{dx^2} \)
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Answer
Now, we differentiate ( \frac{dy}{dx} ) to find ( \frac{d^2y}{dx^2} ):
[ \frac{dy}{dx} = 9x^2 - 16x ]
Differentiating again:
[ \frac{d^2y}{dx^2} = 18x - 16 ]
Step 3
Verify that C has a stationary point when \( x = 2 \)
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Answer
To verify if there is a stationary point at ( x = 2 ), substitute into ( \frac{dy}{dx} ):
[ \frac{dy}{dx} = 9(2)^2 - 16(2) = 36 - 32 = 4 ]
Since ( \frac{dy}{dx} = 0 ) is not satisfied (it's 4), this indicates that there is no stationary point at ( x = 2 ). Hence, more assessments at nearby values may be required.
Step 4
Determine the nature of this stationary point, giving a reason for your answer.
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Answer
As concluded in part (b), since the first derivative ( \frac{dy}{dx} ) at ( x = 2 ) does not equal zero, hence, there is no stationary point at that value. Therefore, there can be no nature to determine since the derivative does not yield a stationary value.