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A. find an equation for the line perpendicular to the tangent to the curve yequals=x cubed minus 9 x plus 7x3−9x+7 at the point ​(33​,77​).
b. what is the smallest slope on the​ curve? at what point on the curve does the curve have this​ slope?
c. find equations for the tangents to the curve at the points where the slope of the curve is 1818.

Respuesta :

A. y=x³-9x+7; y'=3x²-9. Slope at (3,7) [I have assumed that (33,77) is meant to be (3,7)] = 3×9-9=18.
The perpendicular to the tangent has a slope of -1/18 and the equation is (y-7)=-(x-3)/18.
y=-x/18+1/6+7=-x/18+43/6.
B. When is 3x²-9 smallest? When 3x²-9=0, so x=±√3 and y=3√3-9√3+7=7-6√3 and y=7+6√3.
Points are (√3,7-6√3), (-√3,7+6√3).
C. I assume the slope is 18. 3x²-9=18, 3x²=27, x²=9, x=±3, and y=7. The equations are y-7=18(x-3) and y-7=18(x+3), so y=18x-47 and y=18x+61. These lines are parallel to one another.

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