Thread: find 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)

Originally Posted by dat1611

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

Already this is incorrect

ok that was a stupid mistake so now i get

(-1,-16), (1/3,-17.2)

correct!