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"What value of $b$ and $c$ for the curve $y = x^3 + bx^2 + cx$ give the tangent equation $y = 3x - 4$ in the point $(1, -1)$?" This is what I do: The derivative of the function is $3x^2 + 2bx + c$ In the point where $x = 1$, it becomes $3 + 2b + c$. Since the tangent equation is $3x - 4$, $3 + 2b + c = 3$. This means that $2b = -c$. The answer in the book says that $c = -4$, and $b = 2$. However, shouldn't any value that conforms to $2b = -c$ work? I mean, they all give the slope of 3. |
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The question asks you to verify the equation of a tangent line to a given curve at a given point. That point must lie on the curve! You need $x=1$ and $y=-1$ to be a solution to the equation $y=x^3+bx^2+c$, i.e. you need $-1=1+b+c$ or equivalently: $b+c=-2$. You can then do what you did and find the derivative and get the condition that $2b=-c$. Solving the equation $b+c=-2$ and $2b=-c$ simultaneously gives $b=2$ and $c=-4$ as the book says. |
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You have to also make the original function pass through the point $(1,-1)$. That gives the equation $$-1 = 1 + b + c,$$ which implies $b+c = -2$. Given $c = -2b$, as you have shown, this implies $-2 = b+c = b-2b = -b$, so indeed $b = 2$ and hence $c = -4$. |
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$c = -10, b = 5$ fulfills $2b = -c$, but $(1,-1)$ is not a solution to $y = x^3 + 5x^2 - 10x$ Remember, the line and the curve still have to meet at $(1, -1)$ |
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you need to use the fact that $y=-1$ which means that $1 + b + c = -1$ |
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