There exists one and only one straight line which is tangent to the curve
š¦ = š„^4 ā 9š„^2 at two points. determine the x coordinates of these two points.
no, those two points are where the graph crosses the x-axis. One has a negative slope and the other a positive. No way could a single line be tangent at both of them.
y' = 4x^3 - 18x = 2x(2x^2-9)
y'=0 at x = Ā±3/ā2
Since y is an even function, y(a) = y(-a) so the horizontal line through (-3/ā2,-81/4) and (3/ā2,-81/4) is tangent at both points.
Not only that, you can be sure it is the only such point, since y' is an odd function, so y'(-a) = -y'(a) and there are no other pairs of points with the same value.
so would (-3,0) and (3,0) count or would that be considered as a secant line?
To find the x-coordinates of the two points where the straight line is tangent to the curve, we need to solve the equation for the points of tangency.
Step 1: Find the derivative of the curve.
The derivative of the function š¦ = š„^4 ā 9š„^2 can be found by using the power rule for differentiation. The power rule states that if š¦ = š„^n, then its derivative is š¦' = nš„^(n-1).
Differentiating the given equation, we get:
š¦' = 4š„^3 - 18š„
Step 2: Set the derivative equal to the slope of the tangent line.
Since the tangent line is a straight line, its slope remains constant at both points of tangency. Therefore, we need to find the x-values where the derivative of the curve equals the slope of the tangent line.
Let the slope of the tangent line be 'm'. We'll set š¦' = m and solve for š„:
4š„^3 - 18š„ = m
Step 3: Solve for the x-coordinates.
Solving the above equation will give us the x-coordinates of the points where the line is tangent to the curve.
Please provide the value of 'm' to proceed with the calculations.