# 2. Let f(x) = 0.1(x³ – 3x2 – 9x). Then%3Df'(x) = 0.1(3x? – 6x – 9) = 0.3(x + 1)(x – 3)%3Df"(x) = 0.6(x – 1)%3D(a) Solutions to f'(x) = 0 are x =(b) The function f is increasing on the interval(s)%3D (c) The function f is concave down on the interval(s)is an inflection point on the graph of f.(d)

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with example help_outlineImage Transcriptionclose2. Let f(x) = 0.1(x³ – 3x2 – 9x). Then %3D f'(x) = 0.1(3x? – 6x – 9) = 0.3(x + 1)(x – 3) %3D f"(x) = 0.6(x – 1) %3D (a) Solutions to f'(x) = 0 are x = (b) The function f is increasing on the interval(s) %3D fullscreen help_outlineImage Transcriptionclose(c) The function f is concave down on the interval(s) is an inflection point on the graph of f. (d) fullscreen
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### Calculus 