Science:Math Exam Resources/Courses/MATH104/December 2013/Question 01 (e)

MATH104 December 2013

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Question 01 (e)
Find the slope of the tangent line to $\displaystyle y=x^{3}+3x^{2}+2$ at its point of inflection.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
This is a multi-step question. How are inflection points related to the second derivative?

Hint 2
Inflection points may occur where the second derivative is zero.

Hint 3
The slope of the tangent line at the inflection point is the value of the first derivative at that point.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the hint, we begin differentiating. $\displaystyle {\begin{aligned}y&=x^{3}+3x^{2}+2\\{\frac {dy}{dx}}&=3x^{2}+6x\\{\frac {d^{2}y}{dx^{2}}}&=6x+6\end{aligned}}$ Solving for $\displaystyle {\frac {d^{2}y}{dx^{2}}}=0$ gives $\displaystyle 0=6x+6$ and so $\displaystyle x=-1$. A quick check shows that the sign of the second derivative changes at this point and so we know that $\displaystyle x=-1$ is indeed an inflection point. Plugging $x=-1$ into the first derivative gives us the slope of the tangent line at the inflection point: $\displaystyle \left.{\frac {dy}{dx}}\right|_{x=-1}=3(-1)^{2}+6(-1)=3-6={\color {blue}-3}.$

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