Derivative : solution exercise 4 (Intermediate math level)
Given the equation of a curve f(x) = 2x3 - 6x + 5, find out the points on the graph where
the tangent lines are parallel to the bisector of the first quadrant angle. You must also figure out the equation of the
tangent lines and trace the function of the curve and those tangent lines on an XY coordinate grid.
f(x) = 2x3 - 6x + 5
f '(x) = 6x2 - 6
The bisector of the first quadrant angle is the line y = x that divides the first quadrant angle into two 45 degrees angles.
The slope of that straight line is 1. Thus any straight line parallel to the bisector of the first
quadrant angle will have, in its equation y = ax + b, a slope a = 1.
Now we know that the tangent lines will be determined by the equation y = 1x + b. But what is the value of b ?
We want to figure out the x and y-coordinate of the point (x1 ; y1) that lies on the curve f(x) = 2x3 - 6x + 5 and where the tangent line
has a slope = 1. Thus f '(x1) = 1, which means that the derivative of the curve at x = x1 is 1. Let's calculate the value of x1:
We see that it exists only two points on the curve where the tangent lines are parallel to the bisector of the first quadrant angle.
And those two points are:
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The general cartesian equation of those tangent lines is y = 1x + b. |
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y = x + b 8,96 = -1,08 + b b = 10,04 The tangent line to the curve f(x) |
y = x + b 1,0395 = 1,08 + b b = - 0,04 The tangent line to the curve f(x) |
The following graph shows the function f(x) and both tangent lines parallel to the bisector of the first quadrant angle :
How can you trace the graph of an equation ? Using our free graphing tool : the Graph Plotter is an online graphing calculator.