CONTINUOUS FUNCTION WITH F' & GRAPH  CALCULUS
Calculus

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draw a possible graph of a continuous function y=f(x) that satisfies the three conditions: I. f'(x)>0 for 1<x<3 II. f'(x)<0 for x<1 and x>3 III. f'(x)=0 for x=1 and x=3
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What you need to do is draw what's called a "sign chart." Draw a straight, horizontal line and label it as f'(x) from x= ∞ to x= +∞. Then mark the midpoint on the line as x= 0 and mark x= 1 and x = 3 on the right. On top of the line, put a minus sign from ∞ to 1, a plus sign from 1 to 3, and from 3 to +∞; these are the signs of the derivative f'(x), which is the rate of change of f(x). For positive values of f'(x), the graph of f(x) is increasing. For negative values of f'(x), the graph of f(x) is decreasing. For f'(x) = 0, the graph of f(x) is neither increasing nor decreasing; its tangent line is horizontal (parallel with the xaxis). This happens at x= 1 and x= 3 (even though the graph of f(x) is increasing at all xvalues between 1 and 3 and all xvalues after 3, it is not increasing or decreasing at x= 1 or x=3).
Lastly, the values of f(x) don't matter since you're only given the sign of f'(x); the actual yvalues of f(x) don't matter, just the change in y with respect to infinitely small change in x (dy/dx). You can draw the graph between any yvalues that you want.
Ask for clarification if necessary!Oh, and the graph is continuous, so it can't have any discontinuities or "kinks" in it; it must be smooth and have all points connected. The graph of f(x) should look somewhat like an S turned counterclockwise 90°.
Correction! The graph is decreasing for all xvalues greater than 3 because f'(x>3) is negative. Disregard the second part of the last parentheses text in the first reply.
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