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In interval notation, the solution is the set (—°°, 2). Solve 5 - 3* <5x+ 2. Answer1<x[Divide both sides by 8.] In interval notation, the solution is the set (|,°°). Solve -7<2x + 5<9. Answer— 6<x<2[Divide by 2.] In interval notation, the solution is the set (—6,2). Solve 3<4x-l<5. Answer 1 s x < \ [Divide by 4.]
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point on a graph where the concavity of the curve changes (from concave down to concave up, or vice versa) is called a point of inflection (Definition 4.14). By implication (think about what separates positive and negative numbers on a number line), if a point (c, f (c)) is a point of inflection, then f ′ ′ ( c ) = 0 .
MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2.0 (fall 2009) This is a self contained set of lecture notes for Math 221. The notes were written by Sigurd Angenent, starting from an extensive collection of notes and problems compiled by Joel Robbin. The LATEX and Python les
Definition. An inflection point is a point (c, f (c)) on the graph of f where the concavity changes. At such a point, either f ′′(c) = 0 or f ′′(c) does not exist. Procedure for finding the Inflection Points. Step 1. Compute f ′′(x) and determine all points in the domain of f where either f ′′(c) = 0 or f ′′(c) does not exist. Step 2.
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For example, take the function y = x3 +x. dy dx =3x2 +1> 0 for all values of x and d2y dx2 =6x =0 for x =0. This means that there are no stationary points but there is a possible point of inflection at x =0. Since d 2y dx 2 =6x<0 for x<0, and d y dx =6x>0 for x>0 the concavity changes at x =0and so x =0is a point of inflection.
All points of inflection on the graph of must occur either where equals zero or where is undefined. In the following examples, we will use these properties to sketch graphs
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Maxima and minima are also called TURNING points or STATIONARY points. Going from left to right, the gradient is DECREASING up to the point P and then it starts INCREASING again. The point P where the gradient stops decreasing and starts increasing is called an INFLECTION point.
