Find the Critical Points y=x^3-2x^2-4x+1

Problem

y=x3−2*x2−4*x+1

Solution

1. Find the derivative of the function with respect to x to determine the slope of the tangent line.

d(y)/d(x)=3*x2−4*x−4

2. Set the derivative to zero because critical points occur where the derivative is zero or undefined.

3*x2−4*x−4=0

3. Factor the quadratic equation to solve for the values of x

(3*x+2)*(x−2)=0

4. Solve for x by setting each factor equal to zero.

3*x+2=0⇒x=−2/3

x−2=0⇒x=2

5. Find the y-coordinates by substituting the x values back into the original function y=x3−2*x2−4*x+1

y*(−2/3)=(−2/3)3−2*(−2/3)2−4*(−2/3)+1=67/27

y(2)=(2)3−2*(2)2−4*(2)+1=−7

Final Answer

Critical Points:(−2/3,67/27),(2,−7)

Want more problems? Check here!