a curve is such that dy/dx = x^2 - 1
Show that no tagent to this curve is parallel to the line with equation
3x + 2y - 4 = 0
slope of the given line is -3/2
so x^2 - 1 must be equal to -3/2
x^2 - 1 = -3/2
x^2 = 1 - 3/2 = -1/2
BUT a square of a real number cannot be negative, so no solution,
so no such tangent
To determine if any tangent to the curve is parallel to the given line, we need to compare the slopes of the tangent line and the line itself.
Given that the equation of the line is 3x + 2y - 4 = 0, we rearrange it to slope-intercept form: y = (-3/2)x + 2, where the slope is -3/2.
To find the slope of the tangent line to the curve, we differentiate the equation of the curve with respect to x:
dy/dx = x^2 - 1
Now, we want to find values of x that will make dy/dx equal to -3/2, because if the slopes are equal, the lines are parallel.
Set x^2 - 1 equal to -3/2:
x^2 - 1 = -3/2
Multiply both sides by 2 to eliminate the fraction:
2x^2 - 2 = -3
Rearrange the equation:
2x^2 = -3 + 2
2x^2 = -1
Divide both sides by 2:
x^2 = -1/2
The equation cannot be satisfied since the square of a real number cannot be negative. Therefore, there are no values of x that will make the slope of the tangent line equal to -3/2, which means no tangent to the curve is parallel to the line 3x + 2y - 4 = 0.