More+on+Extrema,+inflection+points,+etc

Kevin Wu: a. Point of inflection of the graph of f is possible when the second derivative of f is zero. or when the first derivative of f' is zero, or when f' has horizontal tangent lines, which is given in the problem as x=1 and x=3. b. To find absolute extremas, first find critical numbers f, which is when f' is zero or undefined x=1, 4 Next find endpoints of f x=-1,5 The largest values of these four x values is the absolute maximum value, the smallest is the absolute minimum Absolute maximum: x=5 (area under f' is greatest meaning f is greatest) Absoulte minimum: x=4 (area under f' is smallest meaning f is smallest) c. At x=2, slope=g'=x'f(x)+xf'(x)=0(2)+2(-1)=-2 g(2)=2(6)=12 x=2 y=mx+b 12=-2(2)+b b= 16 y=-2x+16



Julie W. a) relative max when f(x) changes from increasing to decreasing and f'(x) changes from positive to negative. There is a relative max at x=2 so point is (2,2) relative minimum is when f(x) changes from decreasing to increasing and f'(x) changes from negative to positive there is no relative minimum

b) graph

(I made it on paint and it looks very fuzzy compared to what I drew on paper, sorry about that)

c) If g(x) is the integration of f(x) then f(x) is the graph of the derivative, or slope, of g(x). Therefore, the graph of f(x) goes from negative to positive at x=1, so that would be a relative minimum. The graph of f(x) also goes from positive to negative at x=3, which would be a relative maximum.

d) A point of inflection is when f"(x) changes sign, and where f(x) goes from concave up to concave down. If the graph of f(x) is actually the graph of g'(x) then g"(x) is the slope of the graph of f(x). Therefore, the slope of the graph of f(x) changes from positive to negative at x=2, so there is an inflection point at x=2 on the graph of g'(x).