Math exercise and derivative exercises online : How calculate the x,y coordinate of a point ? How calculate the slope (angular coefficient) of a straight line and the slope of a tangent line using mathematics and derivatives ?
Derivative : solution exercise 7 (Intermediate math level exercises)
Find out the coordinates of the points for which the slope of the tangent line to the curve
y = x3 - 12x + 1 is zero.
Given y = f(x) = x3 - 12x + 1
f '(x) = 3x2 - 12
The derivative of f(x) at x = x1 (and y = y1) is equal to the slope of the tangent line to the curve f(x), at (x1 , y1). But that slope must be equal to zero, thus:
f '(x1) = 0 = 3(x1)2 - 12
3(x1)2 = 12
(x1)2 = 12/3 = 4
x1 = +/- 2
y1 = f(x1) = (+/- 2)3 - 12(+/- 2) + 1 = -15 ou 17
The slope of the tangent line to the curve is equal to 0 at the point (2 ; -15) and at the point (-2 ; 17).
Check this on the following graph :
You can clearly see that both tangent lines have a slope equal to zero because the tangent lines are horizontal at x = 2 and at x = -2.
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