Find the Critical Points (x+1)^2(2x-x^2)

Problem

ƒ(x)=(x+1)2*(2*x−x2)

Solution

1. Expand the expression to make differentiation easier.

ƒ(x)=(x2+2*x+1)*(2*x−x2)

ƒ(x)=2*x3−x4+4*x2−2*x3+2*x−x2

ƒ(x)=−x4+3*x2+2*x

2. Differentiate the function with respect to x to find the derivative.

d(ƒ(x))/d(x)=−4*x3+6*x+2

3. Set the derivative to zero to find the critical points.

−4*x3+6*x+2=0

4. Factor the polynomial. Notice that x=−1 is a root because −4*(−1)3+6*(−1)+2=4−6+2=0

(x+1)*(−4*x2+4*x+2)=0

5. Solve for the remaining roots using the quadratic formula on −4*x2+4*x+2=0

x=(−4±√(,4−4*(−4)*(2)))/(2*(−4))

x=(−4±√(,16+32))/(−8)

x=(−4±√(,48))/(−8)

x=(−4±4√(,3))/(−8)

x=(1±√(,3))/2

6. List all values of x where the derivative is zero.

x=−1

x=(1+√(,3))/2

x=(1−√(,3))/2

Final Answer

Critical Points: *x=−1,(1±√(,3))/2

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