Find the Critical Points y=-x^3+x^2+8x+1

Problem

y=−x3+x2+8*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+2*x+8

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

−3*x2+2*x+8=0

3. Multiply by -1 to make the leading coefficient positive, which simplifies factoring.

3*x2−2*x−8=0

4. Factor the quadratic equation by finding two numbers that multiply to −24 and add to −2

(3*x+4)*(x−2)=0

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

3*x+4=0⇒x=−4/3

x−2=0⇒x=2

6. Find the y-coordinates by substituting the x values back into the original function.

y(2)=−(2)3+(2)2+8*(2)+1=−8+4+16+1=13

y*(−4/3)=−(−4/3)3+(−4/3)2+8*(−4/3)+1=64/27+16/9−32/3+1=−143/27

Final Answer

Critical Points:(2,13),(−4/3,−143/27)

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