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Let fx=Ax³+6x-7, where A is a real-valued constant. What is the slope of the line tangent to fx when x=-3 ?

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Final answer:

To find the slope of the tangent line to the curve f(x) at x = -3, the derivative of the function is taken, which is f'(x) = 3Ax^2 + 6, and then we substitute x = -3 to find the specific slope value depending on A.

Step-by-step explanation:

To find the slope of the tangent line to the curve f(x) = Ax^3 + 6x - 7 at x = -3, we must first calculate the derivative of f(x), which gives us the slope of the tangent line at any point x. The derivative f'(x) = 3Ax^2 + 6. Substituting x = -3 into f'(x) gives us the specific slope value at that point: f'(-3) = 3A(-3)^2 + 6 = 27A + 6. Therefore, the slope of the tangent line at x = -3 depends on the value of the constant A.

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