Calculus/Calculus

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Expert: - 1/31/2011


Question
I’m stuck on these Calculus questions.
1.   When a curve attains a maximum value, the slope of its tangent line is equal to what value?
2.   Using your answer to the previous question, find the point at which y=(-3X^2) + 12X attains a maximum value. Your answer will be in the form (a,b) where a and b are numbers.
3.   Test your answer to the previous problem by finding the y values of points that are close to the point you found. (for example, if the point you found was (6,-36) then you might find the y values for x=5 and x=7 and hopefully they’d both be less than -36)
4.   Draw a curve where the slope of the tangent line is always between -1 and 1
5.   Find the equation of the line that is tangent to y=(8x^2) + 2x at the point (1,10)

Thank you for your time.
Elinor

Answer
1. The slope of a smooth curve at the maximum or minimum is 0, so y' = 0.

2. Given y = -3x² + 12x, y' = -6x + 12, so the critical point is at x = 2.
Put x back in to get y, and the point will be at (2,y).

3. Try putting in x = 1.9 and x = 2.1.

4. A curve like this will have a maximum slope of 45° and a minimum slope of -45°.

5. If y = 8x² + 2x, then y' = 16x + 2.
Putting in 1 for x, we can see that y = 10.
Since the value of x at this point is 1, the slope is 16+2 = 18.
Thus, use the point-slope equation to get the line.

That is, y - y0 = m(x - x0) where m is the slope and (x0,y0) is the point.

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