Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:

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Presentation transcript:

Section 4.3b

Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:

Second Derivative Test for Local Extrema Let’s see these with the graphs of the squaring function and the negative squaring function… 1.If and, then has a local maximum at x = c. 2. If and, then has a local minimum at x = c.

Another Quick Example Find the extreme values of Critical Points??? x = 2, –2 Support our answers graphically??? has a local max. at x = –2 has a local min. at x = 2

Let 1. Identify where the extrema of occur. 2. Find the intervals on which is increasing and the intervals on which is decreasing. 3. Find where the graph of is concave up and where it is concave down. 4. Sketch a possible graph for.

Let is continuous since exists. The domain of is all reals, so the domain of is the same. Thus, the critical points of occur only at the zeros of. CP at x = 0, 3 Intervals Sign of Behavior of x < 0 – Dec. 0 < x < 3 – Dec. x > 3 + Inc. Use first derivative test:Local Min. at x = 3 Decreasing on, Increasing on

Let  The zeros of (IP) are at x = 0, 2 Intervals Sign of Behavior of x < 0 + Conc. up 0 < x < 2 – Conc. down x > 2 + Conc. up Concave up on, Concave down on

Let Summarizing info. from both tables: x < 0 Decreasing Concave up 0 < x < 2 Decreasing Concave down 2 < x < 3 Decreasing Concave up x > 3 Increasing Concave up One possibility for the graph of : x = 0 x = 2 x = 3

Let One possibility for the graph of : x = 0 x = 2 x = 3 Note: We are able to recover almost everything about a differentiable function by examining its first derivative… We cannot determine how to place the graph in the x-y plane (vertically)  to position the graph we would need only the value of at one point!!!

A function is continuous on,,, and derivatives have the following properties: x–2 < x < x = 0 does not exist 0 < x < 2 – + x = < x < 4 – – 1. Find where all absolute extrema of occur. 2. Find where the points of inflection of occur. 3. Sketch a possible graph of.

A function is continuous on,,, and derivatives have the following properties: x–2 < x < x = 0 does not exist 0 < x < 2 – + x = < x < 4 – – Absolute Maximum occurs at x = 0 (cannot determine its value) Point of Inflection at x = 2 Absolute Minimum of 1 at x = 4