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A curve has equation $y = x^3 - 48x$ - AQA - A-Level Maths Mechanics - Question 15 - 2018 - Paper 1

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A curve has equation $y = x^3 - 48x$. The point A on the curve has $x$ coordinate $-4$. The point B on the curve has $x$ coordinate $-4 + h$. 15 (a) Show that the... show full transcript

Worked Solution & Example Answer:A curve has equation $y = x^3 - 48x$ - AQA - A-Level Maths Mechanics - Question 15 - 2018 - Paper 1

Step 1

Explain how the result of part (a) can be used to show that A is a stationary point on the curve.

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Answer

To show that point A is a stationary point, we need to analyze the gradient found in part (a).

As h0h \to 0, the gradient of the line AB, which is given by h212hh^2 - 12h, approaches:

limh0(h212h)=0.\lim_{h \to 0}(h^2 - 12h) = 0.

Since the gradient of the curve at point A is the same as the gradient of line AB, we conclude that the gradient approaches zero at point A. Thus, A must be a stationary point on the curve.

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