The domain of f is the set of all real numbers x such that -3<x<5 .
a) For what values of x does f have a relative maximum? Why?
b) For what values of x does f have a relative minimum? Why?
c) On what intervals is the graph of f concave upward? Use f' to justify your answer.
d) Suppose that f(1)=0 . In the xy-plane provided, draw a sketch that shows the general shape of the graph of the function f on the open interval 0<x<2 .
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So where's the graph?
From the graph, severe factors are x = -7, -5, -a million, 3, 5 and 7. the sole community optimal of f is at x = -5, because of the fact f' is beneficial to the left and detrimental to the suited. So it rather is the sole candidate for rather the optimal different than the endpoints. f' is beneficial from -7 to -5, which ability the graph of f will boost from x = -7 to x = -5. as a result, f(-7) < f(-5), and x = -7 won't be able to be an absolute optimal. So now the optimal might desire to be the two x = -5 or x = -7. seem on the "section under the curve" over any era -- this represents the internet substitute in f over that era. case in point, on the era x = (-5, -a million), there's a "internet loss" interior the fee of f, so all of us understand that f(-5) > f(-a million). the part of the region between the curve and the x-axis (easily "above the curve" i assume) represents the quantity that f has decreased on that era. yet now seem on the section under the curves that lie above the x-axis from x = -a million to x = 5, and from x = 5 to x = 7. needless to say, the part of those areas are extra advantageous than the tiny bump under the curve. So, even nevertheless f misplaced some volume on (-5, -a million), it won an excellent extra advantageous volume on (-a million, 7) -- this ability that f(-5) < f(7), and x = 7 is rather the optimal.