My Math Forum The Equation of a Tangent to a Curve

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 August 5th, 2013, 04:05 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 The Equation of a Tangent to a Curve Can anyone please confirm if I have solved this correctly? Many thanks. Q. Show that $f(x)=x^3-6x^2+18x+4$ is an increasing function for all values of $x\in\mathbb{R}$. Attempt: $\frac{dy}{dx}=3x^2-12x+18$ $3(x^2-4x+6)$ $3(x^2-4x+4+2)$ $3((x-2)^2+2)$ Since $(x-2)^2$, $x\in\mathbb{R}$ is always positive, then $3((x-2)^2+2)$ is positive. Therefore, the curve is increasing for all $x\in\mathbb{R}$ as $3((x-2)^2+2)$ is positive for all $x\in\mathbb{R}$.
 August 5th, 2013, 04:34 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 Math Focus: Elementary mathematics and beyond Re: The Equation of a Tangent to a Curve That is correct. What part were you unsure of?
 August 5th, 2013, 05:10 AM #3 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: The Equation of a Tangent to a Curve It was the first question of it's type in the set I'm working with. There are no answers provided in the back of the text book for these questions, for some reason, so I wanted to make sure my first effort was correct before proceeding. Thanks for checking.

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Show that the equation of -x^3 -2x its tangent is always positive

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