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March 26, 2017

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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

  • calculus - ,

    "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!

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