Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 2

If you don't like a sign table:

From Question 2 (a), we know that the critical points of ƒ(x), when ƒ'(x) or does not exist, are at x = ±1, ±√6, and ±√3.

• Testing a value less than -√6, such as -3, we get ƒ'(-3) = 2/3 > 0.
• Testing a point between -√6 and -√3, such as -2, yields ƒ'(-2) = -2 < 0.
• Testing a point between -√3 and -1, such as -1.5, we get ƒ'(-1.5) = -25/3 < 0.
• Finally, we test a point between -1 and 1, such as 0, and get ƒ'(0) = 2/3 > 0.

Normally, we would have to test points between the positive critical points as well, but since our function only has x² terms in it (i.e. ƒ' is even), we know that our negative test points will have identical values to their corresponding positive test points.

(For example, testing -2 and 2 produce identical values of -2, so ƒ(x) < 0 between -√6 and -√3 as well as between √3 and √6.)

Therefore: ƒ(x) is

• increasing on the intervals $(-\infty ,-{\sqrt {6}})\cup (-1,1)\cup ({\sqrt {6}},\infty )$

and

• decreasing on the intervals $(-{\sqrt {6}},-{\sqrt {3}})\cup (-{\sqrt {3}},-1)\cup (1,{\sqrt {3}})\cup ({\sqrt {3}},{\sqrt {6}})$.

Note that the critical points are not included in the intervals of increase and decrease.