i just would like someone to check my work on this one i think ive done it correctly

f(x)= x^3+x^2-x-17

f' = 3x^2-2x-1

3x^2-2x-1=0

(3x+1)(x-1)=0

x= 1, -1/3

1^3+1^2-1-17=-16

(-1/3)^3+(-1/3)^2-(-1/3)-17=-17.3

(1, -16), (-1/3,-17.3)

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- Aug 2nd 2010, 03:03 PMdat1611find the exact values for the horizontal tangent
i just would like someone to check my work on this one i think ive done it correctly

f(x)= x^3+x^2-x-17

f' = 3x^2-2x-1

3x^2-2x-1=0

(3x+1)(x-1)=0

x= 1, -1/3

1^3+1^2-1-17=-16

(-1/3)^3+(-1/3)^2-(-1/3)-17=-17.3

(1, -16), (-1/3,-17.3) - Aug 2nd 2010, 03:09 PMpickslides
- Aug 2nd 2010, 03:22 PMdat1611
ok that was a stupid mistake so now i get

(-1,-16), (1/3,-17.2) - Aug 2nd 2010, 03:41 PMpickslides
correct!