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Can someone help me with this calculus question?

The tangent to the curve y=ax^3+1 at the point where x=2 is inclined at 45 degrees to the x axis.

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  • 8 years ago
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    I'm gonna assume your question is asking you to solve for a? You didn't actually post a question.

    If you want to solve for a, well, let's look at what we know.

    The derivative is the slope of the tangent line at a given point on the curve, right?

    Well, we are given that the slope of the tangent line to the curve at x = 2 is 45 degrees. What is the formula for a line with a slope of 45 degrees?

    Well, that'd just by Y = x + C, where C is any constant number. And we know X is 2, so the derivative of ax^3 + 1 at x = 2 must equal 2 + c.

    The derivative of ax^3 + 1 is going to be 3ax^2, again we know x = 2, so the derivative is 12a = 2 +c. To make our life easy, we can say that C = 0... so a is going to be 2 / 12 or 1/6.

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