Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (e)

MATH104 December 2011

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Question 02 (e)
Consider the function $f(x)={\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}$ Its first and second derivatives are given by $f'(x)={\frac {(x^{2}-1)(x^{2}-6)}{(x^{2}-3)^{2}}}$ and $f''(x)={\frac {2x(x^{2}+9)}{(x^{2}-3)^{3}}}$ Find the coordinates of all local maxima, local minima, and inflection points. Be sure to indicate which is which.

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Hint
If a point is a maximum or minimum, what must be true about its derivative? What can be said about the connection between inflection points and the second derivative?

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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. Local Maxima and Minima To find local maxima and minima, we consider critical points which are points where the derivative is zero or does not exist. In part (c), we got that the critical points were $x=\pm {1},\pm {\sqrt {3}},\pm {\sqrt {6}}.$ In that same part we got that the function was increasing on $(-\infty ,-{\sqrt {6}})\cup (-1,1)\cup ({\sqrt {6}},\infty )$ decreasing on $(-{\sqrt {6}},-{\sqrt {3}})\cup (-{\sqrt {3}},-1)\cup (1,{\sqrt {3}})\cup ({\sqrt {3}},{\sqrt {6}})$. We can immediately exclude -√3 and √3 as potential maximum or minimum because the function is not even defined there. We notice that to the left of x=-√6 the function is increasing and then on the right it is decreasing. Therefore, we have a local maximum at x=-√6. To the left of x=-1, the function is decreasing and then to the right, it is increasing. Therefore x=-1 is a local minimum. To the left of x=1, the function is increasing and to the right it is decreasing. Therefore x=1 is a local maximum. To the left of x=√6 the function is decreasing and to the right it is increasing. Therefore x=√6 is a local minimum. Therefore we conclude that we have two local maxima at $x=-{\sqrt {6}},1$ and two local minima at $x=-1,{\sqrt {6}}$. Inflection Points To find inflection points, we consider where the second derivative is zero or does not exist. In part (d) we got that the second derivative vanished or failed to exist at $x=0,\pm {\sqrt {3}}.$ In that same part we got that the function was concave down on $(-\infty ,-{\sqrt {3}})\cup (0,{\sqrt {3}})$ and concave up on $(-{\sqrt {3}},0)\cup ({\sqrt {3}},\infty )$. Like with the max and min test, we immediately exclude $x=\scriptstyle \pm {\sqrt {3}}$ because these are vertical asymptotes. While the concavity may change on either side of these points, they are not technically defined as inflection points. We see that the only point left to test is x=0. We notice that the function is concave up to the left of x=0 and concave down to the right. Therefore, we conclude that since the concavity changes across this point, x=0, is an inflection point. Exact coordinates We are asked to get the coordinates and we sub the x values for our maxima, minima, and inflection point into the function to find the y values. Therefore we conclude that $\left(-{\sqrt {6}},1-{\frac {4{\sqrt {6}}}{3}}\right)$ is a local maxima, $\left(-1,{\frac {1}{2}}\right)$ is a local minima, $\left(1,{\frac {3}{2}}\right)$ is a local maxima, $\left({\sqrt {6}},1+{\frac {4{\sqrt {6}}}{3}}\right)$ is a local minima, and $\displaystyle {}(0,1)$ is an inflection point.

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