The tangent line to the curve defined by the equation y=(x²+x-2)³+3 at the point (1,3) is confirmed to be tangent at another point on the curve. The derivative dy/dx is calculated as 3(x²+x-2)²(2x+1), which equals zero at x=1, indicating a horizontal tangent line at that point. By solving the equations y=3 and y=(x²+x-2)³+3, two solutions are obtained, confirming the existence of another point of tangency.
PREREQUISITESStudents studying calculus, particularly those focusing on derivatives and tangent lines, as well as educators seeking examples of polynomial tangency in real-world applications.
hallowon said:Homework Statement
show that the tangent line to y=(x2+x-2)3+3 at (1,3) is also tangent to the curve at another point
The Attempt at a Solution
dy/dx=3(x2+x-2)2(2x+1)
at 1
dy/dx = 0
y=mx+b
b=3
y=3