Suppose f(x) is the function 2x^3+ 12x^2+
21x + 3. Find the smallest value of f at which the tangent to the curve is 3.
I did:
f'(x)+ 6x^2+ 24x + 21= 3
= 6(x+3)(x+1), x= -3, -1
the correct answer is supposed to be -8
"Find the smallest value of f at which the tangent to the curve is 3."
Check the values of f(-3) and f(-1) and you'll get the right answer!